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Principal and molar specific heats of gases - class-XI

Description: principal and molar specific heats of gases
Number of Questions: 105
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Tags: physics heat and thermodynamics isothermal and adiabatic processes
Attempted 0/104 Correct 0 Score 0

In an adiabatic change, the pressure $P$ and temperature $T$ of a diatomic gas are related by the relation $P\ \propto \ T^{c}$, where $C$ equals to:

  1. $1.6$

  2. $0.4$

  3. $0.6$

  4. $2.5$


Correct Option: D

The amount of heat necessary to raise the temperature of $0.2 \ mol\ of\ N _2$ at constant pressure from $37^oC$ to $ 337^oC$  will be

  1. $746\ J$

  2. $1746\ J$

  3. $2746\ cal$

  4. $3746\ J$


Correct Option: B
Explanation:

$N _2$ is a diatomic molecule thus its degree of freedom is 5. Its $C _p$ is given as $(1+\displaystyle\dfrac{f}{2})R=(1+\dfrac{5}{2})R=\dfrac{7}{2}R$
Thus, we get the heat required as $Q=nC _p\Delta T=0.2\times \displaystyle\dfrac{7}{2}\times 8.314\times 300=1746  J$

The specific heat of a gas at constant pressure as compared to that at constant volume is

  1. less

  2. equal

  3. more

  4. constant


Correct Option: C
Explanation:

When the gas is heated at constant pressure, some amount of heat is used up in increasing the volume of the gas. For a constant volume process no such heat is required. Thus $C _p>C _v$

The molar specific heat of an ideal gas at constant pressure and volume are $C _p$ and $C _v$ respectively. The value of $C _v$ is

  1. $R$

  2. $\gamma$ R

  3. $\dfrac{R}{\gamma-1}$

  4. $\dfrac{\gamma R}{\gamma-1}$


Correct Option: C
Explanation:

We know that $\displaystyle\dfrac{C _p}{C _v}=\gamma$ and $C _p-C _v=R$.
Thus we get $C _v(\displaystyle\dfrac{C _p}{C _v}-1)=R$
or, $C _v=\displaystyle\dfrac{R}{\gamma -1}$

The gaseous mixture consists of $16\quad $ of helium and $16\quad $ of oxygen. The ratio $\cfrac { { C } _{ p } }{ { C } _{ v } } $ of the mixture is :-

  1. 1.59

  2. 1.62

  3. 1.4

  4. 1.54


Correct Option: B

Calculate the specific heat of a gas at constant volume from the following data. Density of the gas at N.T.P =$19 \times 10 ^ { - 2 } \mathrm { kg } / \mathrm { m } ^ { 3 }$ $\left( C _ { p } / C _ { v } \right)$ = 1.4,J =$4.2 \times 10 ^ { 3 } \mathrm { J } / \mathrm { kcal }$ atmospheric pressure=$1.013 \times 10 ^ { 5 } N / m ^ { 2 }$ (in kcal /kg k)

  1. $2.162$

  2. $1.612$

  3. $1.192$

  4. $2.612$


Correct Option: A

The ratio of the specific heat of air at constant pressure to its specific heat constant volume is

  1. Zero

  2. Greater than one

  3. Less than one

  4. Equal to one


Correct Option: B
Explanation:

The correct answer is option(B).

The ratio of specific heat at constant pressure to the specific heat at constant volume is always greater than one.
As, when the gas is allowed to expand resulting in constant pressure, some of the heat is converted to work resulting in the need of a higher amount of heat to raise the temperature of the gas. Whereas when the volume of the gas is constant, the entire heat supplied is utilized in raising the gas temperature. Hence the heat required the raise the temperature of a unit mass of gas at constant pressure is greater than that required at constant volume. Hence the ratio $c _p:c _v$ is always greater than one.

Which of the following formula is wrong?

  1. $\displaystyle{C _{v} = \dfrac{R}{\gamma - 1}}$

  2. $\displaystyle{C _{p} = \dfrac{\gamma R}{\gamma - 1}}$

  3. $\displaystyle \dfrac{C _{p}}{ C _{v}} = \gamma$

  4. $C _{p} - C _{v} = 2R$


Correct Option: D
Explanation:

The different formula for specific heats is given by:

  • $\dfrac{C _{p}}{C _{v}} = \gamma$
  • $C _{p} - C _{v} = R$
Upon further simplification, we get:
  • $C _{p} = \dfrac{\gamma R}{\gamma -1}$
  • $C _{v} = \dfrac{R}{\gamma -1}$
The incorrect formula is
$C _{p} - C _{v} = 2R$
Hence option D is the answer.

For a gas the ratio of the two specific heats is $\dfrac{5}{3}$. If R $=$ 2 cal /mol-K then the values of $C _{p}$ and $C _{v}$ in cal / mol- K 

  1. $C _p=5 ,C _v=3 $

  2. $C _p=3 ,C _v=4 $

  3. $C _p=4 ,C _v=3 $

  4. $C _p=3 ,C _v=5 $


Correct Option: A
Explanation:

From given data we have $C _{p} - C _{v} = 2 $
and $\dfrac{C _{p}} { C _{v}} = \dfrac {5}{3} $
Solving both gives , 
$C _{p} = 5$ and  $ C _{v} = 3 $

A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. Then $\dfrac{C _{p}}{C _{v}}$ is

  1. 1.67

  2. 2.14

  3. 1.29

  4. 1.33


Correct Option: D
Explanation:

The diatomic molecule has total 6 degress of freedom(3 translational, 2 rotational and 1 vibrational)
Now $C _p$ is given as $(1+\dfrac{f}{2})R=(1+\dfrac{6}{2})R=4R$
and $C _v$ is given as $\dfrac{f}{2}R=\dfrac{6}{2}R=3R$
Thus we get $\dfrac{C _p}{C _v}=\dfrac{4}{3}=1.33$

Which of the following formula is wrong ?

  1. $C _{v}=\dfrac{R}{\gamma -1} $

  2. $\dfrac{C _{p}}{C _{v}}=\gamma $

  3. $C _{p}=\dfrac{\gamma R}{\gamma -1} $

  4. $C _{p}-C _{v}=2R $


Correct Option: D
Explanation:

C$ _p$-C$ _v$ = R and hence option D is incorrect

If the ratio of sp.heat of a gas at constant pressure to that at constant volume is $\gamma $ , the change in internal energy of gas, when the volume changes from V to 2V at constant pressure P is 

  1. $\dfrac{R}{\gamma -1}$

  2. PV

  3. $\dfrac{PV}{\gamma -1}$

  4. $\dfrac{\gamma PV}{\gamma -1}$


Correct Option: C
Explanation:

The change in internal energy in the process should have been,
U = $ nC _v \Delta T $
Now, for this process, if $ \dfrac{{C} _{p}}{{C} _{v}} = \gamma $ and $C _p-C _v=R$
Then, $C _v =  \dfrac{R}{\gamma - 1} $
U = $ \dfrac{nR \Delta T}{\gamma - 1} $
Now, $ nR \Delta T = P(2V - V) $
Thus, U = $ \dfrac{PV}{\gamma - 1} $

In an isobaric process, the correct ratio is :

  1. $\Delta Q:\Delta W=1:1$

  2. $\Delta Q:\Delta W=\gamma :\gamma -1$

  3. $\Delta Q:\Delta W= \gamma -1:\gamma $

  4. $\Delta Q:\Delta W= \gamma :1 $


Correct Option: B
Explanation:

In an isobaric process, pressure is constant.
Heat added in an isobaric process is given by
$\Delta Q=\Delta U+ \Delta W$
$nC _p\Delta T=nC _v\Delta T+nR\Delta T$
$\displaystyle \therefore \dfrac {\Delta Q}{\Delta W}=\dfrac {nC _p\Delta T}{nR\Delta T}=\dfrac {C _p}{R}=\dfrac {\gamma R/\gamma -1}{R}=\dfrac {\gamma }{\gamma -1}$
Option B.

A cylinder of fixed capacity $67.2$ liters contains helium gas at STP. Calculate the amount of heat required to raise the temperature of the gas by $15^{o}C$. ($R=8.314\ J\ mol^{-1}k^{-1}$)

  1. $520\ J$

  2. $560. J$

  3. $620\ J$

  4. $621.2\ J$


Correct Option: B
Explanation:

Since, the process is at constant volume,
Q = U as W = 0
Thus, Q = $ n {C} _{v} \Delta T $
At STP, n = $ \dfrac{PV}{RT} $
Since, He is diatomic, $ {C} _{v} = 2.5R $
Q = $ \dfrac{PV}{RT} \times 2.5R \times 15 $
Substituting the pressure and temperature values at STP, 
P = 1 atm
V = 67.2 L
T = 298 K
we get,
Q = 560.9 J

A diatomic gas is heated at constant pressure. The fraction of the heat energy used to increase the internal energy is 

  1. $ \dfrac{3}{5}$

  2. $ \dfrac{3}{7}$

  3. $ \dfrac {5}{7}$

  4. $ \dfrac {7}{9}$


Correct Option: C
Explanation:

In a diatomic gas, we have $C _p=\dfrac{7}{2}R $ and $C _v=\dfrac{5}{2}R$
The heat is given as $nC _p\Delta T$ and internal energy as $nC _v\Delta T$
Thus we get $\dfrac{U}{Q}$ as $\dfrac{5}{7}$

Four students found set of $C _{p}$ and $C _{v}$[in cal/deg mole] as given below, which of the following set is correct 

  1. $C _{v}=4,C _{p}=2$

  2. $C _{v}=4,C _{p}=3$

  3. $C _{v}=3,C _{p}=4$

  4. $C _{p}=5,C _{v}=3$


Correct Option: D
Explanation:

C$ _{v}$ cannot be greater than $C _{p}$

Hence, option A  and option B are incorrect.

We have the relation $C _{p} - C _{v}$= R ( and its value is 2 cal/mole ) and hence option C is also incorrect.

A solid copper sphere(density $\rho$ and specific heat c) of radius r at an initial temperature $200$K is suspended inside a chamber whose walls are at almost $0$ K. The time required to the temperature of sphere to drop to $100$ K is _________?

  1. $\dfrac{9r\rho c}{72\times 10^6\sigma}$sec.

  2. $\dfrac{7r\rho c}{72\times 10^6\sigma}$sec.

  3. $\dfrac{7r\rho c}{82\times 10^6\sigma}$sec.

  4. $\dfrac{19r\rho c}{72\times 10^7\sigma}$sec.


Correct Option: B

If $C _p$ and $C _v$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M, where R is the molar gas constant:

  1. $C _p - C _v = R/M^2$

  2. $C _p - C _v = R$

  3. $C _p - C _v = R/M$

  4. $C _p - C _v = M/R$


Correct Option: C
Explanation:

By definition 

$dU={ C } _{ v }dT\longrightarrow 1$

also enthalpy,

$H=U+PV\\ or\quad dH=dU+d\left( PV \right) \\ or\quad dH=dU+nRdT\longrightarrow 2$

Also $dH={ C } _{ P }dT\\ \therefore { C } _{ P }dT={ C } _{ V }dT+nRdT\\ \Rightarrow { C } _{ P }={ C } _{ V }+nR\\ or{ C } _{ P }-{ C } _{ V }\quad =nR=\cfrac { Rm }{ M } $

for $m=1$

${ C } _{ P }-{ C } _{ V }=\cfrac { R }{ M } $

${C} _{P}$ and ${C} _{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
${C} _{P}-{C} _{V}=a$ for hydrogen gas
${C} _{P}-{C} _{V}=b$ for nitrogen gas
The correct relation between $a$ and $b$ is then

  1. $a=28b$

  2. $a=\cfrac{1}{14}b$

  3. $a=b$

  4. $a=14b$


Correct Option: D
Explanation:

For ideal gas
${C} _{P}-{C} _{V}=R/M$
If ${C} _{P}$ and ${C} _{V}$ are specific heats $\left( J/kg- _{  }^{ o }{ C } \right) $
$M=$ molar mass of gas
$\Rightarrow a=R/2$ and $b=R/28$
$\Rightarrow$ $a=14b$

A mass of $50g$ of water in a closed vessel with surroundings at a constant temperature takes $2$ minutes to cool from ${30}^{o}C$ to ${25}^{o}C$. A mass of $100g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from ${30}^{o}C$ to ${25}^{o}C$. The specific heat of the liquid is : (The water equivalent of the vessel is $30g$)

  1. $2.0kcal/kg$

  2. $7kcal/g$

  3. $3kcal/kg$

  4. $0.5kcal/kg$


Correct Option: D
Explanation:
As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from $30^{\circ} C$ to $25^{\circ} C$) is same 2 minutes.

$\therefore \left(\dfrac{dQ}{dt} \right) _{water} = \left(\dfrac{dQ}{dt} \right) _{liquid}$

Or $\dfrac{(m _w c _w + W) \Delta T _1}{t _1}  = \dfrac{(m _l c _l + W) \Delta T _2}{t _2}$

$\therefore \Delta T _1=\Delta T _2, \, t _1=t _2$

(w = water equivalent of the vessel)

or $m _w c _w = m _l c _l$

$\therefore$ specific heat of liquid,

$C _l = \dfrac{m _w c _w}{m _l} = \dfrac{50 \times 1}{100} = 0.5 kcal/kg$

Thermal efficiency $=$ .........................   or
$\displaystyle \frac{Heat  Utilised}{Heat  Produced}$

  1. $\displaystyle \frac{Q _4}{Q _T}$

  2. $Q _4 \times Q _T$

  3. $Q _4 + Q _T$

  4. $Q _4 - Q _T$


Correct Option: A

For hydrogen gas $C _{p}-C _{v}=a$ and for Oxygen gas $C _{p}-C _{v}=b $, where $C _{p}$ and $C _{v}$ are molar specific heats. Then the relation between a and b. is

  1. a $=$ 16b

  2. b $=$ 16a

  3. a $=$ 14b

  4. a $=$ b


Correct Option: D
Explanation:

For any ideal gas,$C _p-C _v=nR$, where $R$ is the gas constant.
That is $C _p-C _v$ per mole for any gas is a constant value.
So, $a=b$
Option D.

Three perfect gases at absolute temperatures ${T} _{1},{T} _{2}$ and ${T} _{3}$ are mixed. The masses of molecules are ${m} _{1},{m} _{2}$ and ${m} _{3}$ and the number of molecules are ${n} _{1},{n} _{2}$ and ${n} _{3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

  1. $\cfrac { { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } }{ { n } _{ 1 }+{ n } _{ 2 }+{ n } _{ 3 } } $

  2. $\cfrac { { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ { T } _{ 2 } }^{ 2 }+{ n } _{ 3 }{ { T } _{ 3 } }^{ 2 } }{ { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } } $

  3. $\cfrac { { n } _{ 1 }{ { T } _{ 1 } }^{ 2 }+{ n } _{ 2 }{ { T } _{ 2 } }^{ 2 }+{ n } _{ 3 }{ { T } _{ 3 } }^{ 2 } }{ { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } } $

  4. $\cfrac { \left( { T } _{ 1 }+{ T } _{ 2 }+{ T } _{ 3 } \right) }{ 3 } $


Correct Option: A

For hydrogen gas $C _{p} -C _{v} = a$ and for oxygen gas $C _{p} - C _{v}=b$, where $C _{p}$ and $C _{v}$ are molar specific heats. Then the relation between 'a' and 'b' is

  1. $a=16b$

  2. $b=16a$

  3. $a=4b$

  4. $a =b$


Correct Option: A

The specific heat of air at constant pressure is $1.005\ kJ/kg\ K$ and the specific heat of air at constant volume is $0.718\ kJ/kg\ K$ .Find the specific gas constant.

  1. $0.287\ KJ/kg K$

  2. $0.21\ kJ/kg K$

  3. $0.34\ kJ/kg K$

  4. $0.19\ kJ/kg K$


Correct Option: A
Explanation:

Specific gas constant = Specific heat at constant pressure - Specific heat at constant volume

                                     = 1.005 - 0.718
                                     = 0.287 KJ/kgK

The specific heat of Argon at constant volume is $0.3122 kj/kg K$. Find the specific heat of Argon at constant pressure if  $ R$  $=$8.314 kJ/Kmole K. (Molecular weight of argon$=$ $39.95$)

  1. $520.3$

  2. $530.2$

  3. $230.5$

  4. $302.5$


Correct Option: A
Explanation:
Given,
$C _v=0.3122\ kJ/kg.K$
$R=8.314$
$M=39.95$
$C _{p}=?$
We know,

$C _p-C _v=\dfrac{R}{M}$

$C _p-C _v=\dfrac{8.314}{39.95}=0.2081$

$C _p=0.3122+0.2081=0.5203$

$C _p=520.3\ J/kg.K$

Option $\textbf A$ is the correct answer

Four moles of a perfect gas heated to increase its temperature by ${2^ \circ }C$ absorbs heat of 40 cal at constant volume. If the same gas is heated at constant pressure the amount of heat supplied is (R$=$ 2 cal/mol K)

  1. 28 cal

  2. 56 cal

  3. 84 cal

  4. 94 cal


Correct Option: B
Explanation:
Heat supplied at constant volume
$Q _v=nC _v\triangle T$
$40=4\times C _v\times 2$
$C _v=5$ cal/mol.K
$C _p=C _v+R=5+2=7cal/mol.k$
$\Rightarrow $ Heat supplied at constant pressure
$Q _p=nC _p\triangle T=4\times 7\times 2$
$Q _p=56cal$

Eight spherical droplets, each of radius $'r'$ of a liquid of density $'\phi'$ and surface tension $'T'$ coalesce to form one big drop. If $'s'$ in the specific heat of the liquid. Then the rise in the temperature of the liquid.

  1. $\dfrac {2T}{3r \rho s}$

  2. $\dfrac {3T}{r \rho s}$

  3. $\dfrac {3T}{2r \rho s}$

  4. $\dfrac {T}{r \rho s}$


Correct Option: C

The specific heat at constant volume for the monatomic argon is $0.075 \ kcal/kg-K$, whereas its gram molecular specific heat is $C _v \ = 2.98 \ cal/mol/K$. The mass of the argon atom is (Avogadro's number $= 6.02 \times 10^{23}$ molecules/mol)

  1. $6.60 \times 10^{-23} g$

  2. $3.30 \times 10^{-23}g$

  3. $2.20 \times 10^{-23}g$

  4. $13.20 \times 10^{-23}g$


Correct Option: A
Explanation:

Mass of one mole of argon =$\dfrac{gram \  molecular \  specific \  heat}{specific\   heat \  at \  constant \  volume}=\dfrac{2.98\times 10^{-3}}{0.075}=0.039733 \ g$


Thus mass of each argon atom=$\dfrac{0.0397333}{6.02\times 10^{23}}=6.60\times 10^{-23}g$

If the ratio of specific heat of a gas at constant pressure to that at constant volume is $\gamma$, the change in internal energy of the mass of gas, when the volume changes from $V \ to \ 2V$ at constant pressure P, is

  1. $\dfrac{R}{\gamma- 1}$

  2. $PV$

  3. $\dfrac{PV}{\gamma - 1}$

  4. $\dfrac{\gamma PV}{\gamma - 1}$


Correct Option: C
Explanation:

At constant pressure, change in internal energy$ \Delta U=nC _v\Delta T$
Now,$\dfrac{C _p}{C _v}=\gamma$
$\Rightarrow 1$+$\dfrac{R}{C _v}$=$\gamma$
$\Rightarrow C _v=\dfrac{R}{\gamma -1}$
Using Charle's law, final temperature=2\times initial temperatue=2T
Thus,  $ \Delta U=nC _v(2T-T)=nC _vT=\dfrac{nRT}{\gamma -1}=\dfrac{PV}{\gamma -1}$

A vessel of volume $0.2 m^3$ contains hydrogen gas at temperature $300 K$ and pressure $1 \ bar$. Find the heat (in kcal) required to raise the temperature to $400 K$. (The molar heat capacity of hydrogen at constant volume is $5 \ cal/mol K$)

  1. $4$

  2. $2$

  3. $5$

  4. $8$


Correct Option: A
Explanation:

Using the equation $PV=nRT$ we have
$0.2\times 10^5=n\times 8.314\times 300$
Thus we get n as 8 moles.
Now the heat absorbed is given as 
$Q=W+U=nR\Delta T+nC _v\Delta T$
or
$Q=n(1+\frac{3}{2})R\Delta T$
or
$Q=8\times \frac{5}{2}\times 1.987\times 100=4000  kcal$

The specific heat of a gas 

  1. Has only two value CP and Cv

  2. Has a unique value at a given temperature

  3. Can have any value between 0 and $\infty $

  4. Depends upon the mass of the gas


Correct Option: A

A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is

  1. 75%, 25%

  2. 25% 75%

  3. 60%, 40%

  4. 40%, 60%


Correct Option: C
Explanation:

For isobaric expansion of monatomic gas,
Heat supplied, $\Delta Q=nC _p\Delta T=2.5nR\Delta T$,
Internal energy change, $\Delta U=nC _v\Delta T=1.5nR\Delta T$,
External work,$ \Delta W=P\Delta V=nR\Delta T$
So the heat energy is distributed as 3:2 between internal energy and work, i.e. 60%, and 40% respectively.

The density of a polyatomic gas in standard conditions is $0.795 kg/m^3$. The specific heat of the gas at constant volume is

  1. $930:J/kgK$

  2. $1400:J/kgK$

  3. $1120:J/kgK$

  4. $1600:J/kgK$


Correct Option: B
Explanation:

For given polyatomic gas, applying ideal gas equation PM=dRT under standard conditions, where P=pressure=1atm, M=molar mass, d=density=$0.795kg/m^3$, 

R=universal gas constant, 
T=absolute temperature=$273K$

M=$0.795\times 8.31\times \dfrac{273}{100000}=0.018kg=18g$
Hence, molecule is $H _2O\Rightarrow$ degree of freedom =6
$\Rightarrow$ specific heat at constant volume=$C _v=\dfrac{f}{2}R=3R=3\times 8.31\times \dfrac{ 1000}{18}=1400\ J/kgK$

A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is

  1. $75\%$, $25\%$

  2. $25\%$, $75\%$

  3. $60\%$, $40\%$

  4. $40\%$, $60\%$


Correct Option: C
Explanation:

According to the first law of thermodynamics,
Q = U + W
Now, at constant pressure, 
W = $ P \Delta V = nR \Delta T $
U = $ n {C} _{v} \Delta T $
For, a monoatomic gas, $ {C} _{v} = 1.5 R $
Thus, Q = $ 2.5 \ nR \Delta T $


Now, $ \dfrac{U}{Q} = \dfrac{1.5R}{2.5R} $ = 60 %
Similarly, $ \dfrac{W}{Q} = \dfrac{R}{2.5 R} $ = 40 %

The value of $C _p-C _v=1.00:R$ for a gas in state $A$ and $C _p-C _v=1.06:R$ in another state. If $P _A$ and $P _B$ denote the pressure and $T _A$ and $T _B$ denote the temperatures in the two states, then

  1. $P _A=P _B$, $T _A>T _B$

  2. $P _A>P _B$, $T _A=T _B$

  3. $P _A < P _B$, $T _A>T _B$

  4. $P _A=P _B$, $T _A < T _B$


Correct Option: C
Explanation:

Since we know that,
$C _p-C _V=nR$
Therefore, state A contain less number of moles of gas then state B
Hence, Pressure in state A will be less than Pressure in state B 
whereas Temperature in state A will be greater than temperature in state B 
Since,
$P \propto n$
$T \propto 1/n$
Hence,
${ P } _{ A }<{ P } _{ B }$
$T _A>T _B$
option (C)

Five moles of hydrogen gas are heated from $30^\circ C$ to $60^\circ C$ at constant pressure. Heat given to the gas is (given $R=2:cal/mol^\circ C$)

  1. $750:cal$

  2. $630:cal$

  3. $1050:cal$

  4. $1470:cal$


Correct Option: C
Explanation:

At constant pressure, the heat supplied to a gas is the same as its change in enthalpy,
Thus, Q = $ n {C} _{p} \Delta T $
$ {C} _{p} = 3.5 R $
Q = $ 5 \times 7 \times 30 $
$Q = 1050 \ cal$

'n' number of liquids of masses m,2m,3m,4m, .......... having specific heats S, 2S, 3S, 4S, ...... at temperatures t, 2t, 3t, 4t, ........ are mixed. The resultant temperature of the mixture is

  1. $\frac{3n}{2n+1} t$

  2. $\frac{2n(n+1)}{3(2n+1)} t$

  3. $\frac{3n(n+1)}{2(2n+1)} t$

  4. $\frac{3n(n+1)}{(2n+1)} t$


Correct Option: B

The gas is heated at a constant pressure. The fraction of heat supplied used for external work is 

  1. $ \dfrac{1}{\gamma}$

  2. $\displaystyle(1- \dfrac{1}{\gamma})$

  3. $ \gamma -1$

  4. $\displaystyle(1- \dfrac{1}{\gamma^2})$


Correct Option: B
Explanation:

Heat absorbed =$ \Delta Q=nC _p\Delta T=\gamma nC _v\Delta T$
Internal energy change=$\Delta U=nC _v\Delta T$

Work done=$ \Delta Q- \Delta U=(\gamma -1)nC _v\Delta T$
Required fraction$=\dfrac{\gamma -1}{\gamma}=1-\dfrac{1}{\gamma}$

The specific heat at constant volume for monoatomic argon is $0.075 : kcal/kg-K$, whereas its gram molecular specific heat is $C _v = 2.98 \ cal/molK$. The mass of the argon atom is (Avogrado's number $= 6.02 \times 10^{23} $ molecules/mol)

  1. $6.60 \times 10^{-23} : g$

  2. $3.30 \times 10^{-23} : g$

  3. $2.20 \times 10^{-23} : g$

  4. $13.20 \times 10^{-23} : g$


Correct Option: A
Explanation:

Mass of one mole of argon atoms=Gram molecular specific heat/Specific Heat

$=\dfrac{2.98g}{0.075 \ mol}=39.733\ g/mol$
Thus mass of one atom of argon=$\dfrac{39.733}{6.02\times 10^{23}}g$
$=6.60\times 10^{-23}g$
Hence correct answer is option A.

The mass of a gas molecule can be computed from the specific heat at constant volume. $C _v$ for argon is $0.075:kcal/kg K$. The molecular weight of an argon atom is $(R=2:cal/mol K)$.

  1. $40:kg$

  2. $40\times 10^{-3}:kg$

  3. $20:kg$

  4. $20\times 10^{-3}:kg$


Correct Option: B
Explanation:

$C _v=75\ cal/kgK=75\times M _o cal/molK=\dfrac{3}{2}R$

$\implies 75M _o=\dfrac{3}{2}\times2kg$
$\implies M _o=0.04kg=40\times 10^{-3}kg$

The specific heats of argon at constant pressure and constant volume are $525:J/Kg$ and $315:J/Kg$, respectively. Its density at NTP will be   

  1. $1.77:kg/m^3$

  2. $0.77:kg/m^3$

  3. $1.77:g/m^3$

  4. $0.77:g/m^3$


Correct Option: A
Explanation:

$m(C _P-C _V)T=PV$

$\implies C _P-C _V=\dfrac{PV}{mT}=\dfrac{P}{dT}$
$\implies d=\dfrac{P}{(C _P-C _V)T}=\dfrac{1.01\times 10^{5}}{273\times (525-315)}=1.77\ kg/m^3$

A monoatomic gas expands at a constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is 

  1. $75\%, 25\%$

  2. $25\%, 75\%$

  3. $60\%, 40\%$

  4. $40\%, 60\%$


Correct Option: D
Explanation:

According to the first law of thermodynamics,
Q = U + W
Now, at constant pressure, 
W = $ P \Delta V = nR \Delta T $
U = $ n {C} _{v} \Delta T $
For, a monoatomic gas, $ {C} _{v} = 1.5 R $
Thus, Q = $ 2.5 nR \Delta T $
Now, $ \dfrac{U}{Q} = \dfrac{1.5R}{2.5R}  = 60 \%$
Similarly, $ \dfrac{W}{Q} = \dfrac{R}{2.5 R} = 40 \%$

If for hydrogen $C _p-C _v=m$ and for nitrogen $C _p-C _v=n$, where $C _p$ and $C _v$ refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between $m$ and $n$ is (molecular weight of hydrogen$=2$ and molecular weight of nitrogen$=14$)

  1. $n=14m$

  2. $n=7m$

  3. $m=7n$

  4. $m=14n$


Correct Option: C
Explanation:

For hydrogen, $C _P-C _V=\dfrac{1}{M _{H _2}}\dfrac{dQ}{dT}=m$

For nitrogen, $C _P-C _V=\dfrac{1}{M _{N _2}}\dfrac{dQ}{dT}=n$
$\implies \dfrac{m}{n}=\dfrac{M _{N _2}}{M _{H _2}}=\dfrac{14}{2}=7$

The average degree of freedom per molecule for a gas are $6$. The gas performs $25 J$ of work when it expands at a constant pressure. The heat absorbed by gas is 

  1. $75 \ J$

  2. $100 \ J$

  3. $150\ J$

  4. $125 \ J$


Correct Option: B
Explanation:

For a gas with 'n' degrees of freedom:
$\gamma = 1 + \dfrac{2}{n} = 1 + \dfrac{2}{6} = \dfrac{4}{3}$
$C _{p} = \dfrac{\gamma R}{\gamma - 1} = 4R$
$C _{v} = \dfrac{R}{\gamma - 1} = 3R$

Heat supplied for constant pressure process is $nC _{p}\Delta T$

Change in internal energy $nC _{v} \Delta T$
$\dfrac{\Delta U}{Q} = \dfrac{C _{v}}{C _{p}} = \dfrac{1}{\gamma} = \dfrac{3}{4}$

Hence $\dfrac{W}{Q} = 1 - \dfrac{\Delta U}{Q} = \dfrac{1}{4}$
$\dfrac{W}{Q}=1-\dfrac{3}{4}=\dfrac{1}{4}$
$\implies Q = 100J$

What is the ratio of specific heats of constant pressure and constant volume for $NH _3$

  1. 1.33

  2. 1.44

  3. 1.28

  4. 1.67


Correct Option: C

A reversible adiabatic path on a P- V diagram foran ideal gas passes through state A where P = 0.7$\times $ ${ 10 }^{ 2  }$ N/${ m }^{ -2 }$ and v=0.0049 $ { m }^{ 3  }$, The ratio of specific heat of the gas is 1.4 , The slop of patch at A is:

  1. $2.0 \times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$

  2. $1.0 \times{ 10 }^{ 3\quad }{ Nm }^{ -8}$

  3. $-2.0\times{ 10 }^{ 7\quad }{ Nm }^{ -3 }$

  4. $-1.0\times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$


Correct Option: A

The value of the ratio ${C} _{p}/{C} _{v}$ for hydrogen is $1.67$ a $30K$ but decreases to $1.4$ at $300K$ as more degrees of freedom become active. During this rise in temperature (assume H2 as ideal gas),

  1. ${C} _{p}$ remains constant but ${C} _{v}$ increases

  2. ${C} _{p}$ decreases but ${C} _{v}$ increases

  3. Both ${C} _{p}$ and ${C} _{v}$ decreases by the same amount

  4. Both ${C} _{p}$ and ${C} _{v}$ increase by the same amount


Correct Option: C

A polyatomic gas with six degrees of freedom does $25\ J$ of work when it is expanded at constant pressure. The heat given to the gas is

  1. $100\ J$

  2. $150\ J$

  3. $200\ J$

  4. $250\ J$


Correct Option: A
Explanation:

Degree of freedom, $f=6$
$\Rightarrow C _v=\dfrac{fR}{2}=3R$
$\Rightarrow C _p=C _v+R=4R$
Also work done $=\Delta W$=25J


Thus for isobaric process applying first law,
Heat given($\Delta Q$) $=$ internal energy change$(\Delta U)+\Delta W$
$\Rightarrow nC _p\Delta T=nC _v\Delta T+\Delta W$
$\Rightarrow 4nR\Delta T=3nR\Delta T +25J$
$\Rightarrow nR\Delta T=25J$
Hence, $ \Delta Q=4\times 25=100J$

A gas expands against a constant external pressure of  $2.00 atm, $ increasing its volume by $ 3.40 L.$   Simultaneously, the system absorbs  $400 J $ of heat from its surroundings. What is  $ \Delta E ,$  in joules, for this gas?

  1. $- 689$

  2. $-289$

  3. $+400$

  4. $+289$


Correct Option: B

Consider a classroom that is roughly  $5 { m } \times 10  { m } \times 3  { m }.$  Initially   ${ t } = 20 ^ { \circ }  { C }$  and  $ { P } = 1$ atm. There are  $50$  people in an insulated class loosing energy to the room at the average rate of  $150$  watt per person. How long can they remain in class if the body temperature is  $37 ^ { \circ } \mathrm { C }$  and person feels uncomfortable above this temperature. Molar heat capacity of air  $= ( 7 / 2 ) R.$

  1. $4.34$ minutes

  2. $5.73$ minutes

  3. $6.86$ minutes

  4. $7.79$ minutes


Correct Option: A

Some student find the value of $C _v$ and $C _P$ for two mole of  gas in calorie/gm -mol K.Which pair is most correct?

  1. $C _v=3, C _p=5$

  2. $C _v=3, C _p=6$

  3. $C _v=3, C _p=2$

  4. $C _v=3, C _p=4.2$


Correct Option: A

Assertion : $C _P$ is always greater than $C _V$ in gases.
Reason : Work done at constant pressure is more than at constant volume.

  1. If both assertion and reason are true and reason is the correct explanation of assertion

  2. If both assertion and reason are true but reason is not the correct explanation of assertion

  3. If assertion is true but reason is false

  4. If both assertion and reason are false


Correct Option: A

$C _{P}$ and $C _{V}$ are specific heats at constant pressure and constant volume, respectively. It is observed that $C _{P} - C _{V} = a$ for hydrogen gas $C _{P} - C _{V} = b$ for nitrogen gas. The correct relation between $a$ and $b$ is

  1. $a = b$

  2. $a = 14b$

  3. $a = 28b$

  4. $a = \dfrac {1}{14}b$


Correct Option: B

If $C _{p} and C _{v}$ denoto the specific heats of nitron per unit mass at constant pressure and constant volume rest then 

  1. $C _{p} and C _{v}$=R/28

  2. $C _{p} and C _{v}$=R/14

  3. $C _{p} and C _{v}$=R

  4. $C _{p} and C _{v}$=28R


Correct Option: A
Explanation:

According to Mayer's relation $C _{p}-C _{v}= R/m$

$C _{p}-C _{v}=\dfrac{R}{m}$
for nitrogen $m=28$
$ \therefore C _{p}- C _{v^{2}} R/28$

$C _v,$ respectively, If $\gamma =\dfrac { { C } _{ p } }{ { C } _{ v } } $ and $R$ is the universal gas constant, then $C _v$ is equal to 

  1. $\gamma ^R$

  2. $\dfrac{1+\gamma }{1-\gamma}$

  3. $\dfrac{R}{(\gamma-1)}$

  4. $\dfrac{(\gamma-1)}{R}$


Correct Option: B

Each molecule of gas has f degree of freedom. The ratio $\dfrac { { C } _{ P } }{ { C } _{ V } } =\gamma $for the gas is 

  1. $1+\dfrac { f }{ 2 } $

  2. $1+\dfrac { 1 }{ f } $

  3. $1+\dfrac { 2 }{ f } $

  4. $\dfrac { f }{ 2 } $


Correct Option: A

The molar specific heat at constant  pressure of an ideal gas is ( 7/2) R. the ratio of specific heat at constant pressure to that at constant volume is 

  1. 9/7

  2. 7/5

  3. 8/7

  4. 5/7


Correct Option: A

Ration of $C _p$ and $C _v$ depends upon temperatures according to the following relation

  1. $\gamma \propto T$

  2. $\displaystyle \gamma \propto \frac{1}{T}$

  3. $\gamma \propto \sqrt{T}$

  4. $\gamma \propto T^o$


Correct Option: D
Explanation:

$\gamma =\dfrac{C _p}{C _v}$ i.e, ratio of specific heat capacity at constant pressure and specific heat capacity at constant volume. It doesn't depend on temperature, i.e, it is independent of temperature.

Which type of ideal gas will have the largest value for $C _p-C _v?$

  1. Monoatomic

  2. Diatomic

  3. Polyatomic

  4. The value will be the same for all


Correct Option: D

For an ideal gas

  1. $C _p$ is less than $C _v$

  2. $C _p$ is equal to $C _v$

  3. $C _p$ is greater than $C _v$

  4. $C _p=C _v=0$


Correct Option: C
Explanation:

 For an ideal gas, $C _p$ is greater than $C _v$ because when gas is heated at constant volume, whole of the heat supplied is used to increase the temperature only but when gas is heated at constant pressure, the heat supplied is used to increases both temperature and the volume of gas (heat is used to do work)

The correct option is C.

Adiabatic exponent of a gas is equal to 

  1. $C _p\times C _v$

  2. $\dfrac{C _p}{C _v}$

  3. $C _p-C _v$

  4. $C _p+C _v$


Correct Option: B

The molar specific heat capacity varies as $C=C _v + \beta V$ ($\beta$ is a constant). Then the equation of the process for an ideal gas is given as

  1. $T^{\frac{\beta}{RV} }= constant$

  2. $V^{\frac{\beta T}{R}}=constant$

  3. $T^{-\frac{R}{\beta V}}=constant$

  4. $V^{\frac{R}{\beta T}}=constant$


Correct Option: B

$1$ $\mathrm { g }$ of a steam at $100 ^ { \circ } \mathrm { C }$ melts how much ice at $\mathrm { CC }$ (Latent heat of ice $= 80$ cal/gm and latent heat of steam $ = 540 \mathrm { cal/gm }$



  1. $1 gm$

  2. $2gm$

  3. $4 gm$

  4. $8 gm$


Correct Option: D

The temperature of 5  mole of a gas which was held at constant volume was change from ${ 100 }^{ 0 }$ C to $120^{ 0 }$ C the change in internal energy was found to be 80 joules the total heat capacity of the gas at constant volume will be equal to 

  1. 8 J/K

  2. 0.8 J/K

  3. 4.0 J/K

  4. 0.4 J/K


Correct Option: C

When $1\ mole$ of a monoatomic gas expands at constant pressure the ratio of the heat supplied that increases the internal energy of the gas and that used in expansion is

  1. $\dfrac{2}{3}$

  2. $\dfrac{3}{2}$

  3. $0$

  4. $\infty$


Correct Option: B
Explanation:

Heat supplied to one mole of gas in a constant pressure process is given by: $Q = C _{p}\Delta T$
Change in the internal energy of gas is given by:$\Delta U = C _{v}\Delta T$


The ratio of heat that goes into increasing the internal energy is:
$\dfrac{\Delta U}{Q} = \dfrac{C _{v}}{C _{p}} = \dfrac{1}{\gamma}$

For a mono atomic gas $\gamma = \dfrac{5}{3}$
So, $\dfrac{3}{5}$ ratio of heat goes into increasing the internal energy, and the rest goes into expansion work = $\dfrac{2}{5}$ of heat supplied

Hence, the ratio of heat supplied to increase internal energy by heat supplied to do expansion is $ = \dfrac{3}{2}$

One mole of helium is heated at $0^o$C and constant pressure. How much heat is required to increase its volume threefold?

  1. $3820\ cal$

  2. $382\ cal$

  3. $38.2\ cal$

  4. $3.28\ cal$


Correct Option: A
Explanation:

As it is a constant pressure process, using Charles law we get $\displaystyle\frac{V}{T}=constant$. Thus for a threefold increase in volume we get threefold increase in temperature. Thus we get the final temperature as $3(273)=819 K.$ Thus $\Delta T=819-273=546 K$.
Now as helium is a diatomic molecule, its degree of freedom f is 5. Ths we get $C _p$ for it as $(1+\displaystyle\frac{f}{2})R=(1+\frac{5}{2})R=\frac{7}{2}R$.
Thus heat transferred will be given as $\Delta Q=nC _p\Delta T$
or
$\Delta Q=1(\displaystyle\frac{7}{2})(1.987)(546)=3820\  cal$

When an ideal diatomic gas is heated at constant pressure then what fraction of heat given is used to increase internal energy of gas ? 

  1. $\dfrac{2}{5}$

  2. $\dfrac{3}{5}$

  3. $\dfrac{3}{7}$

  4. $\dfrac{5}{7}$


Correct Option: D
Explanation:

For a diatomic gas we have the degree of freedom as 5. 

Thus heat given at constant pressure is given as $nC _p\Delta T=n(1+\displaystyle\dfrac{5}{2})R\Delta T=n\dfrac{7}{2}R\Delta T$. 
The heat given to change the internal energy is $nC _v\Delta T=n\displaystyle\dfrac{5}{2}R\Delta T$. 
The fraction of internal energy thus used is $\dfrac{5}{7}$ 

One mole of a monoatomic gas and one mole of a diatomic gas are mixed together. What is the molar specific heat at constant volume for the mixture ?

  1. $\dfrac{5}{2} R$

  2. $2 R$

  3. $\dfrac{3}{2} R$

  4. $3 R$


Correct Option: B
Explanation:

$C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here $f$ is the degree of freedom. For monoatomic gas $f=3$ and for diatomic gas $f=5$. 

Thus we get $C _v$ for the mixture as $\displaystyle\dfrac{n _1}{n _1+n _2}(\dfrac{3}{2}R+\dfrac{5}{2}R)=2R$. Here $n _1$ and $n _2$ both are 1.

If water at ${ 0 }^{ \circ  }C.$kept in a container with an open top , is placed in a large evacuated chamber- 

  1. All the water will sported

  2. All the water will French .

  3. Part of the water will vaporize will be formed and reached . equilibrium at the triple point.

  4. ice , water and vapour will be formed and reach equilibrium at the triple points .


Correct Option: A

Equal volumes of monoatomic and diatomic gases of same initial temperature and pressure are mixed. The ratio of the specific heats of the mixture ($C _p/C _v$) will be

  1. $1.53$

  2. $1.52$

  3. $1.5$

  4. $1$


Correct Option: C
Explanation:

$C _p$ for a gas is given as $(1+\displaystyle\dfrac{f}{2})R$ and $C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here f is the degree of freedom. 


For monoatomic gas it is 3 and for a diatomic gas it is 5. 

Thus, we get $C _p$ for the mixture as $\displaystyle\dfrac{5}{2}R+\dfrac{7}{2}R=6R$ and $C _v$ is given as $\displaystyle\dfrac{3}{2}R+\dfrac{5}{2}R=4R$.

Thus, the ratio $\displaystyle\dfrac{C _p}{C _v}$ is given as $\displaystyle\dfrac{3}{2}=1.5$

For an ideal gas during an adiabatic process $\left ( \frac{T^{1}}{P^{2}} \right )^{\frac{1}{5}}$  = constant.  The molar heat capacity at constant volume of the gas is 

  1. 2.5 R

  2. 0.5 R

  3. 3 R

  4. $\frac{7}{2}R$


Correct Option: A

Find the ratio of specific heat at constant pressure to the specific heat at constant volume for ${ NH } _{ 3 }$

  1. 1.33

  2. 1.44

  3. 1.28

  4. 1.67


Correct Option: C

An ideal gas has molar specific heat 5R/2 at constant pressure. If 300 J of heat is given to two moles of gas at constant pressure, the changes in temperature is : 

  1. $ 7.22^oC$

  2. $8.94^oC$

  3. Zero

  4. $5^oC$


Correct Option: D

The volume of 1 kg of hydrogen gas at N.T.P. is ${ 11.2 }m^{ 3 }$. Specific heat of hydrogen at constant volume is $100.46J\quad Kg^{ -1 }{ K }^{ -1 }$.Find the specific heat at constant pressure in $Jkg^{ -1 }{ K }^{ -1 }$?

  1. 120.2

  2. 142.2

  3. 163.4

  4. 182.3


Correct Option: A

The quantity of heat (in J) required to raise the temperature of $1.0\, kg$ of ethanol from $293.45\, K$ to the boiling point and then change the liquid to vapor at that temperature is closest to 
[Given : Boiling point of ethanol $351.45\, K$
              Specific heat capacity of liquid ethanol $2.44\, J\, g^{-1}\, K^{-1}$
               Latent heat of vaporization of ethanol $855\, J \, g^{-1}$]

  1. $1.42\, \times 10^2$

  2. $9.97\, \times 10^2$

  3. $1.42\, \times 10^5$

  4. $9.97\, \times 10^5$


Correct Option: D

A real gas behaves like an ideal gas at which pressure (P) nd temperature (T)?

  1. low P,high T

  2. high P,high T

  3. low P ,low T

  4. high P, low T


Correct Option: D

An ideal monatomic gas follows a law, $P\propto { T }^{ 2 }$ in addition to ideal gas law. Then molar heat capacity for the process is 

  1. $R$

  2. $\dfrac { R }{ 2 } $

  3. $\dfrac { 3R }{ 2 } $

  4. $2R$


Correct Option: A

An ideal gas expands into a vacuum in a rigid vessel. As a result there is :

  1. a change in entropy

  2. a increase of pressure

  3. a change in temperature

  4. a decrease of internal energy


Correct Option: B,C

Which of the following statements are incorrect?
I. If $Q > 0$, heat is added to the system.
II. If $W > 0$, work is done by the system.
III. If $W = 0$, work is done by the system.

  1. II and III

  2. I, II and III

  3. I and II

  4. I and III


Correct Option: A

A monatomic ideal gas expands at constant pressure, with heat Q supplied. The fraction of Q which goes as work done by gas is

  1. 1

  2. $\displaystyle{\dfrac{2}{3}}$

  3. $\displaystyle{\dfrac{3}{5}}$

  4. $\displaystyle{\dfrac{2}{5}}$


Correct Option: D
Explanation:

$Q = nC _p \Delta T$ and $W = P\Delta V = nR\Delta T$
monatomic gas, $\displaystyle{C _p = \dfrac{5R}{2}}$.
$\Rightarrow$$\displaystyle{\dfrac{W}{Q} = \dfrac{2}{5}}$

For a solid with a small expansion coefficient

  1. $ C _p - C _v = R $

  2. $ C _p = C _v $

  3. $ C _p $ is slightly greater than $ C _v $

  4. $ C _p $ is slightly less than $ C _v $


Correct Option: C

When water is heated from $0^{\circ}C$ to $4^{\circ}C$ and $C _{p}$ and $C _{v}$ are its specific heated at constant pressure and constant volume respectively, then:

  1. $C _{p} >C _{v}$

  2. $C _{p}< C _{v}$

  3. $C _{p}=C _{v}$

  4. $C _{p}-C _{v}=R$


Correct Option: B
Explanation:

Water has highest density at $4^{\circ}C$. This changes its properties from other simple fluids.

When water is heated from $0^{\circ}C$ to $4^{\circ}C$, the volume of liquid decreases.
Thus for this transition, $P\Delta V$ is negative.
$\int C _PdT=\int C _VdT+P\Delta V$
$\implies C _P<C _V$

Two moles of ideal helium gas are in a rubber balloon at $30^{o}C$. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35^{o}C$. The amount of heat required in raising the temperature is nearly $($take $R=8.31 J/ mo 1.K)$

  1. $62 J$

  2. $104 J$

  3. $124 J$

  4. $208 J$


Correct Option: D
Explanation:

For isobaric process.
$ \Delta Q= n C _{p} \Delta T$
$=2 \times \dfrac{5}{2} R \times (35-30)$
$= 208 \ J$

The temperature of $5\ moles$ of a gas which was held at constant volume was changed from $100^{o}C$ to $120^{o}C$. The change in the internal energy of the gas was found to be $80\ J$, the total heat capacity of the gas at constant volume will be equal to

  1. $8\ J/K$

  2. $0.8\ J/K$

  3. $4.0\ J/K$

  4. $0.4\ J/K$


Correct Option: C
Explanation:

$dU = nC _v dT$ or, $ 80 = 5 \times C _v(120 - 100)$
$C _v = 4.0\ J/K$

The value of the ratio $C _p/C _v$ for hydrogen is 1.67 at 30 K but decreases to 1.4 at 300 K as more degrees of freedom become active. During this rise in temperature

  1. $C _p$ remains constant but $C _v$ increases

  2. $C _p$ decreases by $C _v$ increases

  3. both $C _p$ and $C _v$ decreases by the same amount

  4. both $C _p$ and $C _v$ increases by the same amount


Correct Option: D
Explanation:

he value of the ratio $\dfrac{Cp}{Cv} $for hydrogen is 1.67 at 30 K but decreases to 1.4 at 300 K as more degrees of freedom become active. During this rise in temperature both $Cp$ and $Cv$ increases by the same amount
 For an ideal gas, $C _p = C _v + R$. If it is a molecular gas, increasing temperature enables vibrational degrees of freedom, so that $C _v$ increases. Hence $\dfrac{C _p}{C _v} = 1 +\dfrac{ R}{C _v}$ decreases.

If $ {C} _{P}$ and $ {C} _{V}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M then which of the following relations is true ?
(R is the molar gas constant)

  1. ${C} _{P}$ - ${C} _{V} = R$

  2. ${C} _{P}$ - ${C} _{V} = R / M$

  3. ${C} _{P}$ - ${C} _{V} = MR$

  4. ${C} _{P}$ - ${C} _{V}$ = $R /{M}^{2} $


Correct Option: B
Explanation:

Let $Cu$ and $Cp$ be molar specific heats of the ideal gas at a 


constant volume and constant pressure, respectively, then

$C _p=M _{c _p}$ and $C _v=M _{c _v}$

Where $C _p$ and $C _v$ are specific heat (per unit mass)

if $C _p$ and $C _v$ are specific heat (for unit mass) of an ideal gas of molecular weight $M$

then specific heat (At constant P) for $M=MC _p$ and 

then specific heat (At constant V) for $M=MC _v$ 

then, $M _{C _p}-M _{C _v}=R$

$\boxed{C _p-C _v=R/M}$

If heat energy $\Delta $ is supplied to an ideal diatomic gas and the increase in internal energy is $\Delta U$, the ratio of $\Delta U:\Delta Q$ is

  1. $7:5$

  2. $5:7$

  3. $5/2 :7/2$

  4. $3:2$


Correct Option: C
Explanation:

For a diatomic gas, the specific heat at constant pressure $C _p=\frac{7}{2}R$ and the specific heat at constant volume $C _v=\dfrac{5}{2}R$

Thus, $\Delta U=nC _v\Delta T=\dfrac{5}{2}nR\Delta T$ and 
$\Delta Q=nC _p\Delta T=\dfrac{7}{2}nR\Delta T$
Hence, $\Delta U:\Delta Q=5/2:7/2$

$310 J$ of heat is required to raise the temperature of $2$ moles of an ideal gas at constant pressure from $25^0C$ to $35^0C$. The amount of heat energy required to raise the temperature of the gas through the same range at constant volume is

  1. $452J$

  2. $276J$

  3. $144J$

  4. $384J$


Correct Option: C
Explanation:
Heat = moles(no.) $\times C _P \times \triangle T$
$\Rightarrow 310=2\times { C } _{ P }\times 10\quad \quad [35-25=10]\\ \Rightarrow { C } _{ P }=15.5J/molK\\ $
$\therefore { C } _{ P }-{ C } _{ V }=R\\ \Rightarrow { C } _{ V }={ C } _{ P }-R=15.5-8.314\\ \Rightarrow { C } _{ V }=7.186J/molK\\ $
$Q=n{ C } _{ V }\triangle T\\ =2\times 7.186\times 10\\ =143.72J\approx 144J$

$C _p$ and $C _v$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C _p-C _v=a$ for hydrogen gas
$C _p-C _v=b$ for nitrogen gas
The correct relation between a and b is :

  1. $a=28 b$

  2. $a=\dfrac{1}{14}b$

  3. $a=b$

  4. $a=14b$


Correct Option: D
Explanation:
For any gas, $C _p-C _v=R$, which is the gas constant.
Hence be it any gas, hydrogen or nitrogen, its value is same.
Here, for ideal gas, $C _p – C _v = R/M$, where $M$ is the mass of one mole of gas.
Mass of one mole of hydrogen  $M = 2$ g and that of nitrogen  $M = 28$ g 
$\therefore$ $a =C _p -C _v= R/2$  (for hydrogen) 
And $b =C _p - C _v = R/28$  (for nitrogen)
  $\implies  a = 14b$

A gaseous mixture consists of $16\ g$ of helium and $16\ g$ of oxygen, then the ratio $\dfrac { { C } _{ p } }{ { C } _{ v } } $of the mixture is

  1. $1.4$

  2. $1.54$

  3. $1.59$

  4. $1.62$


Correct Option: D
Explanation:

Hellium is monoatomic.

Oxygen is diatomic.
Degree of freedom of He, ${ f } _{ 1 }=$3
Degree of freedom of ${ O } _{ 2 }$, ${ f } _{ 2 }=$5
$\therefore \cfrac { { C } _{ p } }{ { C } _{ v } }$ of mixture.
$\cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { [{ N } _{ 1 }(2+{ f } _{ 1 })+{ N } _{ 2 }(2+{ f } _{ 2 })] }{ { N } _{ 1 }{ f } _{ 1 }+{ N } _{ 2 }{ f } _{ 2 } }$
Putting,${ N } _{ 1 }={ N } _{ 2 }=16\quad gm,\quad$ and $\ { f } _{ 1 }=3,\quad { f } _{ 2 }=5$
we get,$ \cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { 3 }{ 2 } \ =1.62$

When a heat of Q is supplied to one mole of a monatomic gas $\left ( \gamma =5/3 \right )$, the molar heat capacity of the gas at constant volume is

  1. $ \dfrac{3R}{4}$

  2. $ \dfrac{5R}{4}$

  3. $ \dfrac{7R}{4}$

  4. $\dfrac{3R}{2}$


Correct Option: D
Explanation:

Given $\gamma =5/3$

i.e $\displaystyle \dfrac {C _p}{C _v}=\dfrac {5}{3}$
and $C _p-C _v=R$

$\therefore \displaystyle \dfrac {C _v+R}{C _v}=\dfrac {5}{3}$
$\displaystyle 1+\dfrac {R}{C _v}=\dfrac {5}{3}$

$C _v= \dfrac{3R}{2}$
Option D.

The molar specific heat of helium at constant volume is $3\ cal/mol^{o}C$ . Heat energy required to raise the temperature of 1gm helium gas by $1^{o}C$ at constant pressure is :

  1. 1.2 cal

  2. 1.25 cal

  3. 3 cal

  4. 4 cal


Correct Option: B
Explanation:

Heat added for a constant pressure process is,


$dQ=dU+dW$

$nC _p\Delta T=nC _v\Delta T+nR\Delta T$

Given 1 gm of Helium, number of moles$= 1/4 =0.25$

$R=2\ cal/mol-K$

$dQ=0.25[3(1)+2(1)]=0.25(5)=1.25\ cal$

Option B.

When 5 moles of gas is heated from $100^{o}C$ to $120^{o}C$ at constant volume, the change in internal energy is 200 J. The specific heat capacity of the gas is

  1. $5\space Jmol^{-1}K^{-1}$

  2. $4\space Jmol^{-1}K^{-1}$

  3. $2\space Jmol^{-1}K^{-1}$

  4. $1\space Jmol^{-1}K^{-1}$


Correct Option: C
Explanation:

Change in internal energy is given by
$dU=nC _v\Delta T$
$200=5(C _v)20$
$C _v=2J/mol.K$
Option C.

A mass of $50$ g of a certain metal at $150^0C$ is immersed in $100$ g of water at $11^0C.$ The final temperature is $20^0C$. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{-1}K^{-1}$.

  1. $0.682 J g^{-1}K^{-1}$

  2. $582 J g^{-1}K^{-1}$

  3. $0.582 J g^{-1}K^{-1}$

  4. $0.0582 J g^{-1}K^{-1}$


Correct Option: C
Explanation:

Mass if the solid , $m _s=50 g$

Initial temperature of the solid, $t _s=150^0 C=150+273=423 k$
Mass of water, $m _w=100 g$
Temperature of the water, $t _w=11^0 C=11+273=293 k$
According to  principle of calorie meter,
Heat gained by water =Heat lost by the solid 
$\begin{array}{l} \therefore { m _{ w } }{ C _{ w } }\left( { t-{ t _{ w } } } \right) ={ m _{ s } }{ C _{ s } }\left( { { t _{ s } }-t } \right)  \ \Rightarrow 100\times 4.2\left( { 293-284 } \right) =50\times { C _{ s } }\times \left( { 423-293 } \right)  \ \Rightarrow 3780=6500\, { C _{ s } } \ \therefore { C _{ s } }=0.582\, \, J/g\, \, k \end{array}$
Hence, Option $C$ is correct.

$n _{1}$ and $n _{2}$ moles of two ideal gases of the thermodynamics constant $\gamma _{1}$ and $\gamma _{2}$ respectively are mixed. $C _{p}/ C _{v}$ for the mixture is

  1. $\dfrac {\gamma _{1} + \gamma _{2}}{2}$

  2. $\dfrac {n _{1}\gamma _{1} + n _{2}\gamma _{2}}{n _{1} + n _{2}}$

  3. $\dfrac {n _{1}\gamma _{2} + n _{2}\gamma _{1}}{n _{1} + n _{2}}$

  4. $\dfrac {n _{1}\gamma _{1}(\gamma _{2} + 1) + n _{2}\gamma _{2}(\gamma _{1} - 1)}{n _{1}(\gamma _{1} - 1) + n _{2}(\gamma _{1} - 1}$


Correct Option: D

A sphere of density $\rho$, specific heat capacity c and radius r, is hung by a thermally insulated thread in an enclosure which is kept at a temperature slightly lower than that of the sphere. The rate of change of temperature for the sphere depends upon the temperature difference between the sphere and the enclosure, and is proportional to then

  1. $\dfrac{c}{r^3 \rho}$

  2. $\dfrac{r^3 \rho}{c}$

  3. $r^3 \rho c$

  4. $\dfrac{1}{r \rho c}$


Correct Option: D
Explanation:

$P. \dfrac{4}{3} \pi r^3 . c. \dfrac{dT}{dt} = e . 4 \pi r^2 \sigma (T - T _0)$
$\dfrac{dT}{dt} \propto \dfrac{1}{Prc}$

1g of $H _{2}$ gas is heated by $1^{o}C$ at constant pressure. The amount of heat spent in expansion of gas is

  1. $\dfrac{4.155}{4.18}cal$

  2. $\dfrac{4.7}{2.1}cal$

  3. $\dfrac{6.8}{2.2}cal$

  4. $\dfrac{1.26}{1.7}cal$


Correct Option: A
Explanation:

The amount of heat spent in the process will be,
Q = $ nR \Delta T $
Q = $ \dfrac{1}{2} \times 8.314 \times 1 $ J
The same value in calorie will be, Q = $ \dfrac{4.155}{4.185} $ cal

The volume of $1\ kg$ of hydrogen gas at $N.T.P$ is $11.2\ m^{3}$. Specific heat of hydrogen at constant volume is $10046J\ kg^{-1}K^{-1}$. Find the specific heat at constant pressure.

  1. $13.8\ kJ/kg-K$

  2. $14.2\ kJ/kg-K$

  3. $16.4\ kJ/kg-K$

  4. $18.3\ kJ/kg-K$


Correct Option: A
Explanation:

Given that,

Mass of hydrogen $m=1\,kg$

Volume of hydrogen $V=11.2\,{{m}^{3}}$

Specific heat of hydrogen at constant volume ${{C} _{V}}=10046\,JK{{g}^{-1}}{{k}^{-1}}$

 We know that,

N.T.P condition as follows

  $ P=1.01\times {{10}^{5}}\,N/{{m}^{2}} $

 $ T={{25}^{0}}C=298\,K $

 Applying gas equation

 $PV=nRT$

 Putting the values in the above equation

 $ PV=nRT $

$ 1.01\times {{10}^{5}}\times 11.2=1\times R\times 298 $

$ R=3795.97\,J/kgk $

 Now according to the Mayer's law

 ${{C} _{P}}-{{C} _{V}}=R$

 Putting the values in the above equation

 $ {{C} _{P}}-{{C} _{V}}=R $

 $ {{C} _{P}}=3795.97+10046 $

$ {{C} _{P}}=13841.97\,JK{{g}^{-1}}{{k}^{-1}} $

Hence, the specific heat of hydrogen at constant pressure is $13841.97\ J/Kg-k$

Molar heat capacity of an ideal gas whose molar heat capacity at constant is $C _v$ for process $P=2e^{2v}$( where P is pressure of gas and V is volume of gas)

  1. $C _v + \dfrac{R}{1+2V}$

  2. $C _v + \dfrac{R}{2V}$

  3. $C _v + \dfrac{R}{V}$

  4. None of these


Correct Option: A
Explanation:

$\begin{array}{l} By\, u\sin  g\, \, first\, law\, of\, Ther{ { moodynamics } }:- \ dQ=dw+dU \ and,\, also\,  \ dQ=nCdT \ dw=Pdv \ dU=n{ C _{ v } }dT \ Now,\, substituting\, them\, in\, the\, first\, law\, we\, get \ \Rightarrow nCdT=PdV+n{ C _{ n } }dT \ C=\frac { { PdV } }{ { ndT } } +{ C _{ v } } \ To\, find\, \, \frac { { PdV } }{ { ndT } } \, we\, will\, use\, the\, ideal\, gas\, equation \ PV=nRT \ 2V{ e^{ 2V } }=nRT\, \, \, \, \, \, \left[ { \, { { Re } }place\, \, P=2{ e^{ 2v } } } \right]  \ Differentiating\, both\, sides\, with\, respect\, to\, T \ 2\left( { { e^{ 2V } }+2V{ e^{ 2V } } } \right) \frac { { dV } }{ { dT } } =nR \ Now,\, from\, this\, we\, have \ \frac { { PdV } }{ { ndT } } =\frac { R }{ { 1+2v } }  \ So,\, we\, get \ C={ C _{ v } }+\frac { R }{ { 1+2v } }  \ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

For a certain gas the heat capacity at constant pressure is greater than that at constant volume by $29.1 J/K$. How many moles of the gas are there?

  1. $13.5 \ mol $

  2. $9.5 \ mol $

  3. $7.5 \ mol $

  4. $3.5 \ mol $


Correct Option: D
Explanation:

We know that for one mole of gas,


${ C } _{ P }-{ C } _{ V }=8.32J/K$

Hence, for n moles,

$n({ C } _{ P }-{ C } _{ V })=8.32n=29.1$

$n=3.5 mol$

Answer is $3.5 mol$

4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 ${ JK }^{ -1 }{ mol }^{ -1 }$. If the speed of sound in this gas at NTP is 952${ ms }^{ -1 }$, then the heat capacity at constant pressure is (Take gas constant R=8.3${ JK }^{ -1 }{ mol }^{ -1 }$)

  1. $8.5{ JK }^{ -1 }{ mol }^{ -1 }$

  2. $8.0{ JK }^{ -1 }{ mol }^{ -1 }$

  3. $7.5{ JK }^{ -1 }{ mol }^{ -1 }$

  4. $7.0{ JK }^{ -1 }{ mol }^{ -1 }$


Correct Option: B

When an ideal diatomic gas is heated at a constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is

  1. $\dfrac {2}{5}$

  2. $\dfrac {3}{5}$

  3. $\dfrac {3}{7}$

  4. $\dfrac {5}{7}$


Correct Option: D
Explanation:

$Th\quad fraction\quad is\quad \frac { \triangle V }{ \triangle Q } \quad =\quad \frac { { _{ n }{ C } _{ v } }\triangle T }{ { _{ n }{ C } _{ p } }\triangle T } \ \frac { \triangle V }{ \triangle Q } =\frac { { C } _{ V } }{ { C } _{ P } } \quad =\quad \frac { 1 }{ Y } \ as\quad we\quad know\quad y\quad =\quad { C } _{ P }/{ C } _{ V }\ y\quad for\quad diatomatic\quad gas\quad :\quad \ { C } _{ P }\quad of\quad diatometic\quad gas\quad :\quad \frac { 7 }{ 2 } \ { C } _{ V }\quad of\quad diatometic\quad gas\quad :\quad \frac { 5 }{ 2 } \ y\quad =\quad \frac { { C } _{ P } }{ { C } _{ V } } =\frac { 7/2 }{ 5/2 } =\frac { 7 }{ 5 } \ \frac { \triangle V }{ \triangle Q } =\frac { 1 }{ y } =\frac { 1 }{ 7/5 } =\frac { 5 }{ 7 } \quad (D)$

For an ideal gas, the heat capacity at constant pressure is larger than that at constant volume because

  1. positive work is done during expansion of the gas by the external pressure

  2. positive work is done during expansion by the gas against external pressure

  3. positive work is done during expansion by the gas against intermolecular forces of attraction

  4. more collisions occur per unit time when volume is kept constant


Correct Option: B
Explanation:

When  heat  is  supplied  at  constant  volume,  temperature  increases accordingly  to  the  ideal  gas  equation.

$P=\dfrac { nRT }{ V } $

as  V  is  constant  and  T  is  increasing,  pressure  will  also  increase.

Than at constant pressure  as temperature is increase volume increases, resulting in expansion of the gas, resulting in positive work, Hence the heat given is used up for expansion and then to increases the internal energy . The heat capacity at constant pressure is larger.

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