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Gay lussac's law - class-XI

Description: gay lussac's law
Number of Questions: 25
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Tags: states of matter chemistry states of matter: gaseous and liquid states states of matter: gases and liquids
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State True or False.

The standard molar volume of an ideal gas is 22.4 L.

  1. True

  2. False


Correct Option: A
Explanation:

The molar volume of a gas is 22.4 litres at STP (standard temperature and pressure). 

At constant volume, pressure is directly proportional to the temperature, this law is popularly known as :

  1. Charle's law

  2. Boyle's law

  3. Gay Lussac law

  4. Avogadro law


Correct Option: C
Explanation:

According to Gay Lussac's law, pressure is directly proportional to temperature at constant volume.
$P\propto T$

Gay Lussac law gives relation between:

  1. $P\,$and$\,V$

  2. $P\,$and$\,T$

  3. $V\,$and$\,T$

  4. none of these


Correct Option: B
Explanation:

Gay Lussac's law states that pressure is directly proportional to temperature at constant volume.

Which gas law should be used to solve a gas problem in which only the pressure and temperature of a gas change?

  1. Boyle's Law

  2. Ideal Gas Law

  3. Combined Gas Law

  4. Gay-Lussac's Law


Correct Option: D
Explanation:

Gay-Lussac's law states that for a given mass and constant volume of  an ideal gas, the pressure exerted on the sides of container is directly propotional to  absolute temperature.

So, for pressure-temperature graph, we use Gay-Lussac's law.

The ratio of volumes of $H _2$ and $O _2$ liberated on electrolysis of water is

  1. $1 : 2$

  2. $1 : 3$

  3. $2 : 1$

  4. $3 : 1$


Correct Option: C
Explanation:

$H _{2}O \rightarrow H^{+} +OH^{-}$
Cathode  :  4$OH^{-} \rightarrow 2H _{2}O +O _{2}+4e^{-} $
Anode :  4$H^{+} +4e^{-} \rightarrow 2H _{2} $
The ratio of volumes of H$ _{2}$ and O$ _{2}$ is $2 : 1$.

A glass container is sealed with a gas at $0.800$ atm pressure and at $25^o$C. The glass container sustain pressure of $2$ atm. Calculate the temperature to which gas can be heated before bursting the container.

  1. $432^o$C.

  2. $472^o$C.

  3. $372^o$C.

  4. $572^o$C.


Correct Option: B
Explanation:
We have,

 $p _1=0.8$atm$,p _2=2$atm $T _1=298$k

Thus,

According to the Gas equation,

$\dfrac{p _1}{T _1}=\dfrac{p _2}{T _2}$

$\dfrac{0.8}{298}=\dfrac{2}{T _2}$

$T _2=745$k

$=472^0$C

Hence, the correct option is $\text{B}$

In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of ${ N } _{ 2 },{ H } _{ 2 }$ and $ { NH } _{ 3 }$ is $1:3:2$. This illustrates law of :

  1. definite proportions

  2. multiple proportions

  3. reciprocal proportions

  4. gaseous proportions


Correct Option: D
Explanation:

As per the Gay Lussac's law of combining volume of gases, the volumes of gaseous reactants and gaseous products bear a simple whole number ratio with each other if they are measured at same temperature and pressure.
In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of $N _2,  H _2$ and $ NH _3$ is $1:3:2$. This illustrates law of Gay Lussac's law of combining volumes of gases.

The volume of a gas is held constant while its temperature is raised. The pressure of the gas exerts on the walls of the container increases because:

  1. The masses of the molecules increases

  2. Each molecule loses more kinetic energy when it strikes the wall

  3. The molecules collide with the wall with relatively greater momentum

  4. The molecules strikes the walls more often


Correct Option: C,D
Explanation:

Volume is kept constant and temperature increased.
We have relation, $PV=nRT$
$P\propto T$
Volume constant
So, $P\propto T$ pressure increases as temperature increases.
As we know the speed or average molecular speed depend on temperature.
As temperature increases molecules collide with wall with greater force or momentum.
Also temperature increases, molecular vibrations increases. So molecule will strike with wall more often.
So option C and D.
A is incorrect because there is no change in masses of molecule when we increase temperature.

A pre-weighed vessel was filled with oxygen at $NTP$ and weighed. It was then evacuated, filled with $SO {2}$ at the same temperature and pressure and again weighed. The weight of oxygen is _____________.

  1. the same as that of $SO _{2}$

  2. $\dfrac{1}{2}$ that of $SO _{2}$

  3. twice that of $SO _{2}$

  4. $\dfrac{1}{4}$ that of $SO _{2}$


Correct Option: B
Explanation:

At constant temperature and pressure, an equal volume of gas contains an equal number of moles.


$\therefore$ Moles of $O _{2}=$ Moles of $SO _{2}$

$1\ mole\ O _{2}=32\ g$

$1\ mole\ SO _{2}=64\ g$

$\therefore wt$ of $O _2=\dfrac{1}{2}$ that of $SO _{2}$

Option $B$ is correct.

A certain vessel $X$ has water and nitrogen gas at a total pressure of 2 $atm$ at 300 $K$. All the contents of the vessel are transferred to another vessel $Y$ having half the capacity of the vessel $X$.The pressure of ${N} _{2}$ in this vessel was 3.8 $atm$ at 300 $K$.The vessel $Y$ is heated to 320 $K$ and the total pressure observed was 4.32 $atm$. The pressure of ${N} _{2}$ at 320 $K$ is :

[Assume that the volume occupied by the gases in the vessel is equal to the volume of the vessel.]

  1. 4.0

  2. 4.05

  3. 5.05

  4. 1.05


Correct Option: B

Law of definite proportions when expressed in terms of volumes becomes:

  1. Dalton's Law

  2. Berzelius hypothesis

  3. Gay-Lussac's Law

  4. Avogadro's Law


Correct Option: C
Explanation:

Law of definite proportions: It states that a given chemical substance (compound) always contains the same elements combined in a fixed proportion by weight. It is also called the law of constant composition.
The law of combining volumes was given by Gay Lussac in 1808. It is applicable only to a gaseous reaction. According to this law, gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure. So, the law of definite proportions when expressed in terms of volumes becomes 'Gay-Lussac's Law'. 
Avogadro's law states that 'equal volumes of all gases, at the same temperature and pressure, have the same number of molecules'.
According to Berzelius hypothesis, equal volumes of all gases contain an equal number of atoms under similar conditions of temperature and pressure.

For which of the following reactions, is Gay Lussac's law not applicable?

  1. Formation of $HI$ from its constituents

  2. Formation of $NH _3$ from its constituents

  3. Formation of $CO _2$ from its constituents

  4. Formation of $SO _3$ from $SO _2$ and $O _2$


Correct Option: C
Explanation:

Gay Lussac's law not applicable to formation of $CO _{2}$ from its constituents as carbon and oxygen exist in different physical states.

Which one is correct about  Gay-Lussac's law?

  1. V/T $=$ k

  2. P/T $=$ k

  3. PV $=$ k

  4. $P _T=P _1+P _2+P _3$

  5. PT $=$ k


Correct Option: B
Explanation:

According to Gay-Lussac's Law :


$\displaystyle \dfrac {P}{T}= k $ 

At constant volume, the pressure of a given mass of a gas varies directly with the temperature.

$\displaystyle P \propto T $ (V and n constant)

Hence, the correct option is B.

Which one proposes a relationship between the combining volumes of gases with respect to the reactants and gaseous products.

  1. Avogadro's number

  2. $P _1V _1=P _2V _2$

  3. $V _1T _2=V _2T _1$

  4. Dalton's Theory

  5. Gay-Lussac's Law


Correct Option: E
Explanation:

Gay-Lussac's Law proposes a relationship between the combining volumes of gases with respect to the reactants and gaseous products. According to this law, when gases react together to produce gaseous products, the volumes of reactants and products bear a simple whole number ratio with each other, provided volumes are measured at same temperature and pressure.

Gay Lussac's Law of combining volumes is applicable for combustion of carbon.

  1. True

  2. False


Correct Option: B
Explanation:
Gay Lussac's Law of Combining Volumes. Gay Lussac's Law of Combining Volumes states that whengases react, they do so in volumeswhich bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

The specific heat of a bivalent metal is 0.16. The approximate equivalent mass of the metal will be:

  1. 40

  2. 20

  3. 80

  4. 10


Correct Option: B
Explanation:

atomic weight = $\dfrac{6.4}{specific heat}$


atomic weight = 40
equivalent weight = $\dfrac{40}{2}$ (bivalent)

equivalent mass = 20

Equal volumes of different gases at any definite temperature and pressure have:

  1. equal weights

  2. equal masses

  3. equal densities

  4. equal number of moles


Correct Option: D
Explanation:

At equal volume of different gases at any definite temperature and pressure have equal no. of particles.

hence, equal number of moles.

Four flasks of 1 litre capacity each arc separately filled with gases $H _2, He, O _2$ and $O _3$. At the same temperature and pressure the ratio of the number of atoms of these gases present in different flasks would be: 

  1. 1 : 1 : 1 : 1

  2. 2 : 1 : 2 : 3

  3. 1 : 2 : 1 : 3

  4. 3 : 2 : 2 : 1


Correct Option: A
Explanation:

Ans ; A

Here in this question 4 different  types of gasses are filled in same volume of flasks i.e. all 4 types of gasses have same number of molecules.
THe ratio of number of atoms of these gasses present in different flasks would be = 1:1:1:1

The volume occupied by 0.01 moles of helium gas at STP is:

  1. $0.224  l$

  2. $22.4  l$

  3. $2240  l$

  4. $2.24  l$


Correct Option: A
Explanation:

As 1 mole of a gaseous substance occupies 22.4 litre ($22.4 l =$ gram molar volume)
Volume occupied $=GMV \times 0.01$( gram molar volume )

Hence, volume occupied $= 22.4\times0.01$
                                            $= 0.224l$

A mixture of CO and $CO _2$ has a density of $1.5 g/l $ at $27^o$C and $760$ mm pressure. If $1\ l$ of the mixture is exposed to alkali, what would be the pressure of the remaining gas at the same volume and temperature?

  1. $533$ mm

  2. $473$ mm

  3. $335$ mm

  4. $595$ mm


Correct Option: C
Explanation:

The given density$=d=\dfrac{m}{volume}$ …….$(1)$


so the given mass is a mixture of CO & $CO _2$

$m=n _{CO}\times 28+n _{CO _2}\times 44$

${$ using no. of moles$=\dfrac{mass}{mol. mass}}$

Also using $PV=nRT$
$V=\dfrac{(n _{CO}+n _{CO _2})RT}{P}$

Putting in $(1)$
$1.5=\dfrac{n _{CO}\times 28+44\times n _{CO _2}}{{(n _{CO _2}+n _{CO})\times 0.082\times 300}}$ $[{760$mm$=1$atm$}, P=1atm]$

$1.5=\dfrac{28n _{CO}+44n _{CO _2}}{24.6(n _{CO}+n _{CO _2})}$


$\Rightarrow 36.9(n _{CO}+n _{CO _2})=28n _{CO}+44n _{CO _2}$

$9n _{CO}=7n _{CO _2}$

After the reaction with alkali, all $CO _2$ will be used so the remaining pressure will be of $CO$.

$P _{CO _2}=\dfrac{n _{CO}}{n _{CO}+n _{CO _2}}\times P _{Total}$

$=\dfrac{n _{CO}}{n _{CO}+\dfrac{9}{7}n _{CO}}$

$=\dfrac{7}{16}P _{Total}=\dfrac{7}{16}\times 760$

$ \approx 335$ mm
Option C.

Chlorofluorocarbons such as $CCl _3F\ (M=137.5)$ and $CCl _2F _2\ (M=121)$ have been linked to ozone depletion in Antarctica. As of $2004$, these gases were found in $275$ and $605$ parts per trillion $(10^{12})$, ny volume. What are the concentrations of these gases under conditions typical of Antarctica stratosphere ($200\ K$ and $0.08\ atm$)? $(R=0.08\ l-atm/K-mol)$   

  1. $[CCl _3F]=1.375\times 10^{-12}\ mol\ l^{-1}$,
    $[CCl _2F _2]=3.025\times 10^{-12}\ mol\ l^{-1}$

  2. $[CCl _3F]=2.75\times 10^{-14}\ mol\ l^{-1}$,
    $[CCl _2F _2]=6.05\times 10^{-14}\ mol\ l^{-1}$

  3. $[CCl _3F]=2.75\times 10^{-10}\ mol\ l^{-1}$,
    $[CCl _2F _2]=6.05\times 10^{-10}\ mol\ l^{-1}$

  4. $[CCl _3F]=1.375\times 10^{-13}\ mol\ l^{-1}$,
    $[CCl _2F _2]=3.025\times 10^{-12}\ mol\ l^{-1}$


Correct Option: A
Explanation:

$CCl _3F(mw)=133.5$
$CCl _2F _2(mw)=121$
By given data first we calculate density of solution
$\dfrac {P}{S}=\dfrac {RT}{Mw}\quad R=0.08\quad T=200$
$P=0.08$
for $CCl _3F-S(CCl _3F)=\dfrac {0.08\times 133.5}{0.08\times 200}$
$S=0.6875$
for $CCl _2F _2=\dfrac {0.08\times 121}{0.08\times 200}$
$=0.605$
given data in $(PPt)$
$10^{12}\ gm$ solution contains $=175\ gm$
$10^{3}\ gm$ solution contains $=\dfrac {275}{10^{12}}\times 10^3$
Mass $=275\times 10^{-9}\ gm$
find the volume $d=\dfrac {M}{v}$
$v=\dfrac {1000}{0.6785}=1454.5$
Molarity of $CCl _3F=\dfrac {275\times 10^{-9}}{137.5\times 1454.5}$
$=1.37\times 10^{-12}$
Similarly for $CCl _2F _2$
$10^{12}\ gm$ solution certain $=605$
$10^{3}\ gm$ solution certain $=\dfrac {605}{10^{12}}\times 10^3$
Mass $=605\times 10^{-9}$
$d=\dfrac {m}{v}\quad v=\dfrac {m}{v}=\dfrac {1000}{605}$
$v=1652.89$
molarity $=\dfrac {605\times 10^{-9}}{121\times 1652.89}$
$M=3.02\times 10^{-12}\ mol\ l^3$

A quantity of $4$g of oxygen occupies $10$ L at a particular pressure and temperature. If the pressure of gas is doubled and absolute temperature is halved, in order to maintain constant volume.

  1. $3$ g gas should be removed from container

  2. $3$ g gas should be added in the container

  3. $16$ g gas should be added in the container

  4. $12$ g gas should be added in the container


Correct Option: D
Explanation:

Given that,
$m _1=4g$ $V _1=10L$,$P _1=P$,$T _1=T$

$m _2=?$, $V _2=10L$,$P _2=2P$, $T _2=\dfrac{T}{2}$

We know that,

$PV=\dfrac{m}{M}RT$

At constant volume, V

$\dfrac{P _1}{P _2}=\dfrac{m _1RT _1}{m _2RT _2}$

$\dfrac{P}{2P}=\dfrac{4\times T}{m _2\times \dfrac{T}{2}}$


$\dfrac{1}{2}=\dfrac{8}{m _2}$

$m _2=16$g

Mass of gas added into container$=16-4=12$g.

In the reaction $N _{2}+3H _{2}\rightarrow 2NH _{3} $, the ratio by volume of $N _{2},\ H _{2} :$ and$: NH _{3}$ is $1 : 3 : 2$. 


This illustrates the law of:

  1. definite proportion

  2. multiple proportion

  3. reciprocal proportion

  4. gaseous volumes


Correct Option: D
Explanation:

In the reaction, $N _{2}+3H _{2}\rightarrow 2NH _{3} $, the ratio by volume of $N _{2},\ H _{2} $ and $ NH _{3}$ is $1 : 3 : 2$. This illustrates the law of Gaseous volumes or Gay Lussac's law of combining volumes of gases.

According to this law, when gases react together to produce gaseous products, the volume of reactants and products bear a simple whole-number ratio with each other, provided volumes are measured at the same temperature and pressure.

So, the correct option is $D$.

To use Gay-Lussac's Law, which of the following needs to remain constant?

  1. Volume and the number of moles of a gas

  2. Pressure and temperature

  3. Temperature and the number of moles of a gas

  4. Pressure and the number of moles of a gas

  5. Temperature and volume


Correct Option: A
Explanation:

$P \propto T $
At constant volume and number of moles of a gas, the pressure is directly proportional to the absolute temperature.

Hence, option $A$ is correct.

A closed vessel contains equal number of nitrogen and oxygen molecules at a pressure of $p\ mm$. If nitrogen is removed from the system then the pressure will be:

  1. $p$

  2. $2p$

  3. $p/2$

  4. $p^2$


Correct Option: C
Explanation:
At constant $V, P\propto n$

$\therefore$ if no. of moles is halved then pressure will also reduce to $p/2$.

Option C is correct.
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