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Work done by an ideal gas in isothermal expansion - class-XI

Description: work done by an ideal gas in isothermal expansion
Number of Questions: 39
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Tags: thermodynamics isothermal and adiabatic processes heat and thermodynamics physics
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An ideal gas has initial volume V and pressure P. In doubling its volume the minimum work done will be in the following process(of given processes)

  1. Isobaric process

  2. Isothermal process

  3. Adiabatic process

  4. None of the above.


Correct Option: C

Two difference gases of molecular masses $M _1$ and $M _2$ are at the same temperature. What is the ratio of their mean square speeds?

  1. $\dfrac{M _1}{M _2}$

  2. $\dfrac{M _2}{M _1}$

  3. $\sqrt {\dfrac{M _1}{M _2}}$

  4. $\sqrt {\dfrac{M _2}{M _1}}$


Correct Option: A
Explanation:

Mean squared speed $=\cfrac{3RT}{M}$

$\cfrac{V _1}{V _2}=\cfrac{M _1}{M _2}$

A diatomic gas which has initial volume of $10$ litre is isothermally compressed to $1/15^{th}$ of its original volume where initial pressure is $10^5$ Pascal. If temperature is $27^o$C then find the work done by gas.

  1. $-2.70\times 10^3$J

  2. $2.70\times 10^3$J

  3. $-1.35\times 10^3$J

  4. $1.35\times 10^3$J


Correct Option: A
Explanation:

$w=nRT ln\left(\dfrac{v _2}{v _1}\right)$
$w=P _0V _0ln\left(\dfrac{v _2}{v _1}\right)$
$w=10^5\times 10\times 10^{-3}ln\left(\dfrac{1}{15}\right)$
$w=-2.70\times 10^3J$.

One mole of an ideal gas undergoes an isothermal change at temperature T so that its volume V is doubled. R is the molar gas constant. Work done by the gas during this change is :

  1. RT $\ln 4$

  2. RT $\ln 3$

  3. RT $ \ln 2$

  4. RT $ \ln 1$


Correct Option: C
Explanation:

Under isothermal process work is given by the relation $W = RT \ln (\dfrac{V _{f}}{V _{i}})$
Therefore work will be $W = RT \ln(2)$

The slope of adiabatic curve is ________ than the slope of an isothermal curve.

  1. Greater.

  2. lesser

  3. data insufficient

  4. can be both a and b


Correct Option: A

Three moles of an ideal gas $\left (C _{P} = \dfrac {7R}{2}\right )$ at pressure $P _{A}$ and temperature $T _{A}$ is isothermally expanded to twice the initial volume. The gas is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure $P _{A}$.
Calculate the net work done by the gas and the net heat supplied to the gas during the complete process.

  1. $0.579\ RT _{A}, \triangle Q = 0.579\ RT _{A}$.

  2. $79\ RT _{A}, \triangle Q = 0.679\ RT _{A}$.

  3. $0.9\ RT _{A}, \triangle Q = 0.779\ RT _{A}$.

  4. $0.7\ RT _{A}, \triangle Q = 0.979\ RT _{A}$.


Correct Option: A

Two soap bubbles having radii $3\ cm$ and $4\ cm$ in vacuum, coalesce under isothermal conditions. The radius of the new bubble is

  1. $1\ cm$

  2. $5\ cm$

  3. $7\ cm$

  4. $3.5\ cm$


Correct Option: B

One mole of an ideal gas u ndergoes a process:
$P = \dfrac{P _0}{1+(V _0/ V)^2}$.
Here $P _0$ and $V _0$ are constants. change in temperature of the gas when volume is changed from $V=V _0$ to $V = 2V _0$ is: 

  1. $-\dfrac{2P _0V _0}{5R}$

  2. $\dfrac{11P _0V _0}{10R}$

  3. $-\dfrac{5P _0V _0}{4R}$

  4. $P _0V _0$


Correct Option: B

Let $Q$ and $W$ denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.

  1. $Q = 0$

  2. $W = 0$

  3. $Q \neq W$

  4. $Q = W$


Correct Option: D

Work done during isothermal expansion of one mole of an ideal gas $10$ atm to $1$ atm at $300\ K$ is

  1. $-4938.8\ J$

  2. $4938.8\ J$

  3. $-5744\ J$

  4. $6257.2\ J$


Correct Option: C

Identical cylinders contain helium at 2.5 atm and agron at 1 atm respectively. If the are filled  in one of the cylinder,the pressure would be.

  1. 3.6 atm

  2. 1.75 atm

  3. 1.5 atm

  4. 1.0 atm


Correct Option: B
Explanation:

Given that,

Pressure ${{p} _{1}}=2.5\,atm$

Pressure ${{P} _{2}}=1\,atm$

Volume ${{V} _{1}}={{V} _{2}}=V$

Both the cylinders are similar, volume of both the gases is equal

Let the volume of gases be V

The amount of pressure P in one of the cylinder will be equal to the total pressure at equilibrium

Now,

  $ P=\dfrac{{{P} _{1}}{{V} _{1}}+{{P} _{2}}{{V} _{2}}}{{{V} _{1}}+{{V} _{2}}} $

 $ P=\dfrac{2.5\times V+1\times V}{2V} $

 $ P=1.75\ atm $

Hence, the pressure is $1.75\ atm$

One mole of an ideal monoatomic gas is at $360K$ and a pressure of $10 ^ { 5 } Pa.$ It is compressed at constant pressure until its volume is halved. Taking $R$ as $8.3 J{ mol } ^ { - 1 }{ K } ^ { - 1 }$ and the initial volume of the gas as $3.0 \times 10 ^ { - 2 } { m } ^ { 3 }$ , the work done on the gas is

  1. $-1500 J$

  2. $+1500 J$

  3. $-3000 J$

  4. $+3000 J$


Correct Option: A

An ideal gas is taken from state $A$ (pressure $P$, volume $V$) to state $B$ (pressure $\displaystyle\frac{P}{2}$, volume $2V$) along a straight line path in the pressure-volume diagram. Select the correct statements from the following.

  1. The work done by the gas in the process $A$ to $B$ exceeds the work done that would be done by it if the system were taken from $A$ to $B$ along an isotherm.

  2. In the temperature-volume diagram, the path $AB$ becomes a part of a parabola.

  3. In the pressure-temperature diagram, the path $AB$ becomes a part of hyperbola.

  4. In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases.


Correct Option: A,B,D
Explanation:

The given process follows a linear P-V relation given by: $ \dfrac{p-P}{v-V}=\dfrac{P-P/2}{V-2V}=-P/2V$
$\Rightarrow p-P=\dfrac{-P}{2V}(v-V) \Rightarrow p=-\dfrac{Pv}{2V}+3P/2$        .......(i)


Here work done $\Delta W$=area under P-V diagram =$3PV/4=0.75PV$
For isotherm, pv=PV $\Rightarrow$ work done $\Delta W$=$PV\ln2=0.693PV$

Hence statement a is correct
For t-v diagram, replace p in (i) by $\dfrac{nRt}{v}$
Hence, $t=-\dfrac{Pv^2}{2nRV}+\dfrac{3Pv}{2nR}$

The above eqn. has t equalling a quadratic in v. 
Hence, t-v diagram is a parabola $\Rightarrow$ statement b is correct.

Clearly the above parabola is concave downwards , hence has a maxima.
The t has same initial and final value implying the maxima occurs in-between.

Hence statement d is correct.

A fixed mass of a gas is first heated isobarically to double the volume and then cooled isochorically to decrease the temperature back to the initial value. By what factor would the work done by the gas decreased, had the process been isothermal?

  1. $2$

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

  3. $\ln 2$

  4. $\ln 3$


Correct Option: C
Explanation:

Let initial pressure and volume be P  and V respectively.
Then after isobaric expansion they are P and 2V. Here work done=$P\Delta V=P(2V-V)=PV$
To bring to initial temperature new pressure=$\dfrac{PV}{2V}=P/2$. 


So after isochoric process they are $P/2  \ and \  2V$ . Here, $\Delta V=0\Rightarrow$ work done=$P\Delta V=0$
Hence, total work done in the 2 successive processes=PV +0=PV
For isothermal expansion, work done=$PVln(\dfrac{V _f}{V _i})=PV\ln2$
Hence required factor=$\dfrac{PV \ln2}{PV}=ln2$

Work done in reversible isothermal process by an ideal gas is given by

  1. $2.303 \text { nRT log } \frac { V _ { 2 } } { V _ { 1 } }$

  2. $\frac { n R } { ( y - 1 ) } \left( T _ { 2 } - T _ { 1 } \right)$

  3. $2.303 \text { nRT log } \frac { V _ { 1 } } { V _ { 2 } }$

  4. None


Correct Option: A

$0\cdot 5\ mole$ of oxygen and $0\cdot 5\ mole$ of nitrogen, each having $V$ and temperature $T$ are mixed isothermally to have the total volume $2\ V$. The maximum work done is :

  1. $RT\ \log _{e}2\ joule$

  2. $\dfrac {RT}{2}\ \log _{e}2\ joule$

  3. $RT\ (\log _{e}4)\ joule$

  4. $zero$


Correct Option: C

A gas expands from $1l$ to $3l$ at atmospheric pressure. The work done by the gas is about 

  1. $2\ J$

  2. $200\ J$

  3. $300\ J$

  4. $2 \times 10^{5}\ J$


Correct Option: B

4 atm pressure to attain 1 atm pressure by result of isothermal  expansion. The work done by the gas during expansion is nearly.

  1. 155 J

  2. 255 J

  3. 355 J

  4. 555 J


Correct Option: D
Explanation:

Given,

$V=1L=0.001m^3$
$P=4atm$
$\dfrac{P _1}{P _2}=4$
From ideal gas,
$PV=nRT$
$T=\dfrac{PV}{nR}$
Work done during the isothermal expansion,
$W=nRTln(\dfrac{P _1}{P _2})$
$W=nR.\dfrac{PV}{nR}ln(4)$
$W=PVln(4)$
$W=4\times 10^5\times 0.001\times ln(4)$
$W=555J$
The correct option is D.

During the process A -B of an ideal gas: 

  1. work done on the gas is zero

  2. density of the gas is constant

  3. slope of line AB from the T - axis is inversely proportional to the number of moles of the gas

  4. slope of line AB from the T - axis is directly proportional to the number of moles of the gas


Correct Option: A

A gas follow a general process as $PV RT + 3V$ for $1\ mole$ of gas. If it expands isobarically till temperature is doubled, then the work done by the gas is (initial temperature and pressure are $T _{0}$ and $P _{0}$ respectively).

  1. $\dfrac {P _{0}T _{0}R}{(2P _{0} - 3)}$

  2. $\dfrac {P _{0}T _{0}R}{(P _{0} - 3)}$

  3. $\dfrac {P _{0}T _{0}R}{(P _{0}V - 3)}$

  4. $\dfrac {3P _{0}V _{0}}{R}$


Correct Option: A

The work done by 100 calorie of heat in isothermal expansion of ideal gas is 

  1. 418.4J

  2. 4.184J

  3. 41.84J

  4. None


Correct Option: A
Explanation:

One calorie is equivalent to $4.184$ joules 

or,   $J=4.18 J/cal$.
Work done (W)=$ = 4.18 \times 500 = 2090\,J$
As $1$ calorie is equal to $4.184$ calorie then $100$ calorie is equal to $ = 4.184 \times 100 = 418.4$

Work done during isothermal expansion depends on change in

  1. volume

  2. pressure

  3. both (a) and (b)

  4. none of these


Correct Option: C
Explanation:

Work done in isothermal expansion is given as $W=nRT\ln(\displaystyle\frac{V _f}{V _i})=nRT\ln(\frac{P _i}{P _f})$. Thus it depends both on pressure and volume.

Expansion of 1 mole of ideal gas is taking place from 2 litres to 8 litres at 
300 K against 1 atm pressure. calculate $ \Delta JK^{-1} mol^{-1} $ given $ R= \dfrac {8.3J}{mol-K}, 1 lit-atm= 100 j,in 2=0.693) $

  1. $11.5$

  2. $13.5$

  3. $9.5$

  4. $22.5$


Correct Option: B

The pressure and volume of a given mass of gas at a given temperature are $ \mathrm{P}  $ and $ \mathrm{V}  $ respectively.Keeping temperature constant, the pressure is increased by 10$  \%  $ and then decreased by 10$  \%  $ .The volume how will be -

  1. less than $ \mathrm{V} $

  2. more than $ \mathrm{V} $

  3. equal to $ \mathrm{V} $

  4. less than $ V $ for diatomic and more than $ V $ for monoatomic


Correct Option: C

Two bulbs of volume V and 4V contains gas at pressures of 5 atm,. 1 atm and at temperatures of 300K and 400K respectively. When these bulbs are joined by narrow tube keeping their temperature at their initial values. The pressure of the system is

  1. 1 atm

  2. 2 atm

  3. 2.5 atm

  4. 3 atm


Correct Option: A

In isobaric process of ideal gas $(f = 5)$ work done by gas is equal to $10\ J$. Then heat given to gas during process is

  1. $25\ J$

  2. $15\ J$

  3. $45\ J$

  4. $35\ J$


Correct Option: D
Explanation:

$\triangle Q = \left (\dfrac {f}{2} + 1\right ) nR\triangle T = \dfrac {7}{2} \times 10 = 35\ J$.

One mole of an ideal gas at $300K$ is expanded isothermally from an initial volume of $1litre$ to $10litres$. The $\Delta E$ for this process is $(R=2cal.mol-1K-1)$

  1. $1381.1cal$

  2. zero

  3. $163.7cal$

  4. $9lit.atm$


Correct Option: B

An Ideal gas undergoes an isobaric process. If its heat capacity is $C _v$ at constant volume and number of mole $n$. then the ratio of work done by gas to heat given to gas when temperature of gas changes by $\Delta T$ is:

  1. $\left(\dfrac{nR}{c _v + R}\right)$

  2. $\left(\dfrac{R}{c _v + R}\right)$

  3. $\left(\dfrac{nR}{c _v - R}\right)$

  4. $\left(\dfrac{R}{c _v - R}\right)$


Correct Option: A
Explanation:

$\dfrac{f}{2} R = \dfrac{C _v}{n}$
$W = nR \Delta T$
$\Delta Q = \left(\dfrac{f}{2} + 1\right) nR \, \Delta T$
$\dfrac{W}{\Delta Q} = \left(\dfrac{2}{f + 2} \right) = \dfrac{2}{\dfrac{2C _v}{nR} + 2} = \left(\dfrac{nR}{C _v + R} \right)$

Three moles of an ideal gas kept at a constant temperature at $300 K$ are compressed from a volume of $4 L$ to $1 L$. The work done in the process is

  1. $-10368 J$

  2. $-110368 J$

  3. $12000 J$

  4. $120368 J$


Correct Option: A
Explanation:

Work done in an isothermal process is given by 
$\displaystyle{W = 2.3026nRT\log _{10}\frac{V _2}{V _1}}$
Here, $n = 3, R = 8.31 J/mol^oC$
$T = 300 K$, $V _1= 4 L$, $V _2 = 1 L$
Hence, $\displaystyle{W = 2.3026 \times 3 \times 8.31 \times 300 \times log _{10}\frac{1}{4}}$
= $17221.15 (-2\log _{10} 2$)
= $-17221.15 \times 2 \times 0.3010 = -10368J$ 

A physical quantity that is conserved in a process

  1. must have the same value for all observers

  2. can never take negative values

  3. must be dimensionless

  4. need not necessarily be a scalar


Correct Option: A

For an isothermal expansion of an ideal gas, mark wrong statement

  1. there is no change in the temperature of the gas

  2. there is no change in the internal energy of the gas

  3. the work done by the gas is equal to the heat supplied to the gas

  4. the work done by the gas is equal to the change in its internal energy


Correct Option: D
Explanation:

$In\quad isothermal\quad expansion.\ The\quad temperature\quad is\quad constant,\ so\quad the\quad change\quad in\quad temperature,\quad \triangle T\quad =\quad 0,\ as\quad \triangle v\quad =\quad \frac { 3 }{ 2 } \quad R& T\ and\quad \triangle T\quad =\quad 0\ \therefore \quad \quad \triangle V\quad =\quad 0\ \ \quad \quad \quad \quad \quad \quad \quad \triangle V\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad 0\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad Q\quad =\quad W\ Work\quad done\quad in\quad isothermal\quad process\quad =\quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \ \quad \quad \quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \quad \neq \quad 0\quad \ \quad \therefore \quad \triangle W\quad \neq \quad \triangle V\ Answer\quad :\quad D$

If a given mass of gas occupies a volume of 10 cc at 1 atmospheric pressure and temperature 100$^o$C. What will be its volume at 4 atmospheric pressure, the temperature being the same?

  1. 100 cc

  2. 400 cc

  3. 1.04 cc

  4. 2.5 cc


Correct Option: D
Explanation:

It is an isothermal process.
$P _1 V _1 = P _2 V _2$
$1 \times 10 = 4 \times V _2$
$V _2 = 2.5 cc$

The cyclic process from X to Y is an isothermal process.
If the pressure of the gas at X is 4.0 kPa, and the volume is 6.0 cubic meters, and if the pressure at Y is 8.0 kPa, what is the volume of the gas at Y?

  1. 12.0 cubic meters

  2. 16.0 cubic meters

  3. 3.0 cubic meters

  4. 4.0 cubic meters

  5. 2.0 cubic meters


Correct Option: C
Explanation:
Given :    $P _x = 4.0$ kPa          $V _x = 6.0$  $m^3$                         $P _y = 8.0$ kPa 
X to Y is an isothermal process, thus temperature remains constant   i.e     $T = constant$  or   $PV = constant$
$\implies$     $P _yV _y = P _xV _x$
OR       $8.0 \times V _y  = 4.0 \times 6.0$                     $\implies V _y = 3.0$  $m^3$

The work done y a gas is an isothermal change where 1 refers to initial state and 2 refers to final state is

  1. $\mu R({T _2} - {T _1})\ln \left( {\frac{{{V _2}}}{{{V _1}}}} \right)$

  2. $\mu R{T _1}\ln \left( {\frac{{{V _2}}}{{{V _1}}}} \right)$

  3. $\mu R{T _2}\ln \left( {\frac{{{V _1}}}{{{V _2}}}} \right)$

  4. $\mu R\left( {\frac{{{T _1} + {T _2}}}{2}} \right)[\ln {V _2} - \ln {V _1}]$


Correct Option: B
Explanation:

$w=-\int PdV   PV=nRT$
$T\rightarrow const$
$=-\int _{v _1}^{v _2}\frac {nRT}{V}dV$
$=\mu RT _1ln(\frac {V _2}{V _1})$

A thin tube of uniform cross-section is sealed at both ends. When it lies horizontally, the middle $5$cm length contains mercury and the two equal ends contain air at the same pressure P. When the tube is held at an angle of $60^o$ with the vertical, then the lengths of the air columns above and below the mercury column are $46$cm and $44.5$cm respectively. Calculate the pressure P in cm of mercury. The temperature of the system is kept at $30^o$C.

  1. $75.4$

  2. $45.8$

  3. $67.5$

  4. $89.3$


Correct Option: A
Explanation:

Let A be the area of cross-section of the tube. When the tube is horizontal, the $5$cm column of Hg is in the middle, so length of air column on either side at pressure $P=\dfrac{46+44.5}{2}=45.25$cm
When the tube is held at $60^o$ with the vertical, the lengths of air columns at the bottom and the top are $44.5$cm and $46$cm respectively. If $P _1$ and $P _2$ are their pressures, then $P _1-P _2=5\cos 60^o=5\times \dfrac{1}{2}=\dfrac{5}{2}$cm of Hg
Using Boyle's law for constant temperature,
$PV=P _1V _1=P _2V _2$
$P\times A\times 45.25=P _1\times A\times 44.5=P _2\times A\times 46$
$\therefore \dfrac{P\times 45.25}{44.5}-\dfrac{P\times 45.25}{46}=\dfrac{5}{2}$
or $P=\dfrac{5\times 44.5\times 46}{2\times 45.25\times 1.5}=75.4$cm of Hg

An ideal gas system undergoes an isothermal process, then the work done during the process is:

  1. $nRT ln\dfrac { { V } _{ 2 }}{ { V } _{ 1 } }$

  2. $nRT ln\dfrac { { V } _{ 1 }}{ { V } _{ 2 }}$

  3. $2nRT ln\dfrac { { V } _{ 2 }}{ { V } _{ 1 }}$

  4. $2nRT ln\dfrac { { V } _{ 1 }}{ { V } _{ 2 }}$


Correct Option: A
Explanation:
Let an ideal gas goes isothermally from its initial state $(P _1V _1)$ to the final state $(P _2V _2)$. Then the work done,

$W=\int _{V _1}^{V _2}P\,dV$

We have 

$PV=nRT$

or

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

Then,

$W=\int _{V _1}^{V _2} \dfrac{nRT}{V} dV=nRT\int _{V _1}^{V _2}\dfrac 1V dV$

$=nRT \,ln(V _2)-ln(V _1)=nRT\,ln\dfrac{V _2}{V _1}$

A vertical cyclinder with heat - conducting with heat conducting walls is closed at the bottom and its fitted with a smooth light piston. It contains one mole of an ideal gas. The temperature of the gas is always equal to the surrounding's temperature $T _o$ . The piston is moved up slowly to increase the volume of the gas to $n$ times. Which of the following is incorrect?

  1. Work done by the gas in $RT _o\ln (n)$.

  2. Work done against the atmosphere is $RT _o(n-1).$

  3. There is no change in the internal energy of the gas.

  4. The final presure of the gas is $\dfrac{1}{n-1}$ times its initial pressure.


Correct Option: D
Explanation:

Work Done by gas (isothermal process) : 

$W=nRT\ln(\dfrac { { V } _{ 2 } }{ { V } _{ 1 } } )=R{ T } _{ o }\ln(n)$
Work Done against Atmosphere (Pressure of atmosphere is constant P & Volume is decreased): 

$W=P\triangle V\quad =P(nV-V)=PV(n-1)=R{ T } _{ o }(n-1)$

As there is no change in temperature hence no change is Internal Energy (Always in synchronization with atmosphere).
 Ideal Gas Equation:
$PV=nRT\ PV=R{ T } _{ o }\ P'(nV)=R{ T } _{ o }\ P'=\dfrac{P}{n}$

Two moles of a gas is expanded to double its volume by two different processors. One is isobaric and the other is isothermal. If ${w} _{1}$ and ${w} _{2}$ are the works done respectively, then

  1. ${ w } _{ 2 }=\cfrac { { w } _{ 1 } }{ \ln { 2 } } $

  2. ${ w } _{ 2 }={ w } _{ 1 }$

  3. ${ w } _{ 2 }={ w } _{ 1 }\ln { 2 } $

  4. ${ w } _{ 1 }^{ 2 }={ w } _{ 2 }\ln { 2 } $


Correct Option: C
Explanation:

Volume is changed from  $V _1 = V _o$  to  $V _2 = 2V _o$.
Temperature and pressure of gas at $V _o$ are  $T _o$ and $P _o$ respectively.
Thus, from ideal gas equation   $P _o V _o =nRT _o$           .....(1)
Work done in isobaric process  $w _1 = P _o (V _2-V _1) = P _o (2V _o - V _o) = P _o V _o$       .....(2)
Work done in isothermal process  $w _2 = nRT _o \ln\dfrac{V _2}{V _1}$
$\therefore$   $w _2 = nRT _o \ln \dfrac{2V _o }{V _o} = nRT _o \ln 2$
Or   $w _2 = P _oV _o \ln 2$   (from 1)
$\implies \ w _2 = w _1\ln 2$  

Let $ \triangle W _a and \triangle W _b $ the work done by the system A and B respectively in the previous question 

  1. $ \triangle W _a > \triangle W _b $

  2. $ \triangle W _a = \triangle _b $

  3. $ \triangle W _a < \triangle W _b $

  4. The relation between $ \triangle W _a and \triangle W _b $ cannot be deduced.


Correct Option: C
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