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Kinetic Theory of Gases

139216 An insulated box containing a diatomic gas of molar mass $M$ is moving with a velocity $v$ . The box is suddenly stopped. The resulting change in temperature is (where, $R$ is the gas constant)

1

\frac{M v^{2}}{2 R}

2

\frac{M v^{2}}{3 R}

3

\frac{{Mv}^{2}}{5 R}

4

\frac{2 {Mv}^{2}}{5 R}

Explanation:

C For diatomic gas, $f = 5$
When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises.
The kinetic energy of gas $= \frac{1}{2} {Mv}^{2} \dots$ ...(i)
Change in internal energy, $Δ U = {nC}_{v} Δ T$
For diatomic gas $C_{v} = \frac{5}{2} R$
From equation, (i) and equation (ii), we get -
${nC}_{v} Δ T = n \frac{1}{2} {Mv}^{2}$
$n \frac{5}{2} R Δ T = n \frac{1}{2} {Mv}^{2}$
$5 R Δ T = {Mv}^{2}$
Change in temperature, $Δ T = \frac{{Mv}^{2}}{5 R}$

CG PET -2018

Kinetic Theory of Gases

139217 The rms speed of oxygen molecule at a certain temperature is $600 {ms}^{- 1}$ . If the temperature is doubled and oxygen molecule dissociates into atomic oxygen atoms, the new rms speed is

1

120 {ms}^{- 1}

2

150 {ms}^{- 1}

3

1200 {ms}^{- 1}

4

600 {ms}^{- 1}

[

Explanation:

C Initial Temperature $= T$
Final Temperature $= 2 T$
Velocity $(v_{rms}) = 600 m / \sec$ .
We know,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$600 = \sqrt{\frac{3 RT}{M}}$
Due to dissociation,
$v_{rms} = \sqrt{\frac{3 R (2 T)}{M / 2}}$
On divide equation (i) by equation (ii), we get-
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 RT}{M}}}{\sqrt{\frac{3 R 2 T}{M / 2}}}$
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{12 RT}{M}}}$
$\frac{600}{v_{rms}} = \frac{1}{2} \Rightarrow v_{rms} = 1200 m / s$
The new rms speed is $1200 m / s$ .

AP EAMCET (23.04.2018) Shift-2

Kinetic Theory of Gases

139218 A thermally insulated vessel with nitrogen gas at $27^{\circ} C$ is moving with a velocity of $100 {ms}^{- 1}$ . If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(Assume entire loss in $KE$ of the gas is given as heat to gas and $R = 8.3 {Jmol}^{- 1} K^{- 1}$ )

1 1.1

2 0.93

3 0.5

4 2.25

Explanation:

D Given that,
Temperature, $T = 27^{\circ} C + 273 = 300 K$
$R = 8.3 J / mol K$
Velocity $= 100 m / \sec$ .
There are $n$ mole of $N_{2} = 28 gm$
$= 28 \times 10^{- 3} kg$
$f = 5$
Increment in pressure due to change of temperature is
$Δ PV = nR Δ T$
$Δ P = \frac{nR Δ T}{V}$
From ideal gas equation -
$PV = nRT$
$P = \frac{nRT}{V}$
Equation (i) divided by equation (ii), we get -
$\frac{Δ P}{P} = \frac{nR Δ T}{nRT} [∵ Δ T = \frac{{Mv}^{2}}{fR}]$
$\frac{Δ P}{P} = \frac{{nMv}^{2}}{fnRT}$
$\frac{Δ P}{P} = \frac{{Mv}^{2}}{fRT}$
So, percentage change in pressure is -
$\frac{Δ P}{P} \times 100 = \frac{28 \times 10^{- 3} \times 100 \times 100 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10}{15 \times 8.3}$
$\frac{Δ P}{P} % = \frac{280}{124.5}$
$\frac{Δ P}{P} % = 2.2489$
$\frac{Δ P}{P} % = 2.25 %$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139221 $V_{1}$ is the speed of sound in a diatomic gas at $273^{\circ} C$ and $V_{2}$ is the r.m.s. speed of its molecules at $273 K$ , then $\frac{V_{1}}{V_{2}}$

1

\sqrt{\frac{15}{14}}

2

\sqrt{\frac{14}{15}}

3

\sqrt{\frac{7}{8}}

4

\sqrt{\frac{8}{7}}

[

Explanation:

B Given, $T_{1} = 273^{\circ} C + 273 K = 546 K$ $T_{2} = 273 K$
$γ_{diatomic} = \frac{7}{5}$
We know,
$V_{1} = \sqrt{\frac{γ {RT}_{1}}{M}} and {(V_{2})}_{rms} = \sqrt{\frac{3 {RT}_{2}}{M}}$
Then,
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{\frac{γ {RT}_{1}}{M}}}{\sqrt{\frac{3 {RT}_{2}}{M}}} = \frac{\sqrt{γ T_{1}}}{\sqrt{3 T_{2}}}$
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{7 \times 546}}{\sqrt{5 \times 3 \times 273}} = \sqrt{\frac{14}{15}}$

Kinetic Theory of Gases

139222 The absolute temperature of a gas is increased to 16 times of the original temperature. The rms speed of its molecules will become

1 4 times

2 16 times

3 64 times

4 256 times

Explanation:

A Let ${(V_{rms})}_{1} = V$
$T_{1} = T$
$T_{2} = 16 T$
$\frac{{(V_{rms})}_{1}}{{(V_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{(V)}{{(V_{rms})}_{2}} = \sqrt{\frac{T}{16 T}} = \frac{1}{4}$
${(V_{rms})}_{2} = 4 V$

J and K-CET-2017

Kinetic Theory of Gases

139216 An insulated box containing a diatomic gas of molar mass $M$ is moving with a velocity $v$ . The box is suddenly stopped. The resulting change in temperature is (where, $R$ is the gas constant)

1

\frac{M v^{2}}{2 R}

2

\frac{M v^{2}}{3 R}

3

\frac{{Mv}^{2}}{5 R}

4

\frac{2 {Mv}^{2}}{5 R}

Explanation:

C For diatomic gas, $f = 5$
When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises.
The kinetic energy of gas $= \frac{1}{2} {Mv}^{2} \dots$ ...(i)
Change in internal energy, $Δ U = {nC}_{v} Δ T$
For diatomic gas $C_{v} = \frac{5}{2} R$
From equation, (i) and equation (ii), we get -
${nC}_{v} Δ T = n \frac{1}{2} {Mv}^{2}$
$n \frac{5}{2} R Δ T = n \frac{1}{2} {Mv}^{2}$
$5 R Δ T = {Mv}^{2}$
Change in temperature, $Δ T = \frac{{Mv}^{2}}{5 R}$

CG PET -2018

Kinetic Theory of Gases

139217 The rms speed of oxygen molecule at a certain temperature is $600 {ms}^{- 1}$ . If the temperature is doubled and oxygen molecule dissociates into atomic oxygen atoms, the new rms speed is

1

120 {ms}^{- 1}

2

150 {ms}^{- 1}

3

1200 {ms}^{- 1}

4

600 {ms}^{- 1}

[

Explanation:

C Initial Temperature $= T$
Final Temperature $= 2 T$
Velocity $(v_{rms}) = 600 m / \sec$ .
We know,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$600 = \sqrt{\frac{3 RT}{M}}$
Due to dissociation,
$v_{rms} = \sqrt{\frac{3 R (2 T)}{M / 2}}$
On divide equation (i) by equation (ii), we get-
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 RT}{M}}}{\sqrt{\frac{3 R 2 T}{M / 2}}}$
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{12 RT}{M}}}$
$\frac{600}{v_{rms}} = \frac{1}{2} \Rightarrow v_{rms} = 1200 m / s$
The new rms speed is $1200 m / s$ .

AP EAMCET (23.04.2018) Shift-2

Kinetic Theory of Gases

139218 A thermally insulated vessel with nitrogen gas at $27^{\circ} C$ is moving with a velocity of $100 {ms}^{- 1}$ . If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(Assume entire loss in $KE$ of the gas is given as heat to gas and $R = 8.3 {Jmol}^{- 1} K^{- 1}$ )

1 1.1

2 0.93

3 0.5

4 2.25

Explanation:

D Given that,
Temperature, $T = 27^{\circ} C + 273 = 300 K$
$R = 8.3 J / mol K$
Velocity $= 100 m / \sec$ .
There are $n$ mole of $N_{2} = 28 gm$
$= 28 \times 10^{- 3} kg$
$f = 5$
Increment in pressure due to change of temperature is
$Δ PV = nR Δ T$
$Δ P = \frac{nR Δ T}{V}$
From ideal gas equation -
$PV = nRT$
$P = \frac{nRT}{V}$
Equation (i) divided by equation (ii), we get -
$\frac{Δ P}{P} = \frac{nR Δ T}{nRT} [∵ Δ T = \frac{{Mv}^{2}}{fR}]$
$\frac{Δ P}{P} = \frac{{nMv}^{2}}{fnRT}$
$\frac{Δ P}{P} = \frac{{Mv}^{2}}{fRT}$
So, percentage change in pressure is -
$\frac{Δ P}{P} \times 100 = \frac{28 \times 10^{- 3} \times 100 \times 100 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10}{15 \times 8.3}$
$\frac{Δ P}{P} % = \frac{280}{124.5}$
$\frac{Δ P}{P} % = 2.2489$
$\frac{Δ P}{P} % = 2.25 %$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139221 $V_{1}$ is the speed of sound in a diatomic gas at $273^{\circ} C$ and $V_{2}$ is the r.m.s. speed of its molecules at $273 K$ , then $\frac{V_{1}}{V_{2}}$

1

\sqrt{\frac{15}{14}}

2

\sqrt{\frac{14}{15}}

3

\sqrt{\frac{7}{8}}

4

\sqrt{\frac{8}{7}}

[

Explanation:

B Given, $T_{1} = 273^{\circ} C + 273 K = 546 K$ $T_{2} = 273 K$
$γ_{diatomic} = \frac{7}{5}$
We know,
$V_{1} = \sqrt{\frac{γ {RT}_{1}}{M}} and {(V_{2})}_{rms} = \sqrt{\frac{3 {RT}_{2}}{M}}$
Then,
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{\frac{γ {RT}_{1}}{M}}}{\sqrt{\frac{3 {RT}_{2}}{M}}} = \frac{\sqrt{γ T_{1}}}{\sqrt{3 T_{2}}}$
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{7 \times 546}}{\sqrt{5 \times 3 \times 273}} = \sqrt{\frac{14}{15}}$

Kinetic Theory of Gases

139222 The absolute temperature of a gas is increased to 16 times of the original temperature. The rms speed of its molecules will become

1 4 times

2 16 times

3 64 times

4 256 times

Explanation:

A Let ${(V_{rms})}_{1} = V$
$T_{1} = T$
$T_{2} = 16 T$
$\frac{{(V_{rms})}_{1}}{{(V_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{(V)}{{(V_{rms})}_{2}} = \sqrt{\frac{T}{16 T}} = \frac{1}{4}$
${(V_{rms})}_{2} = 4 V$

J and K-CET-2017

Kinetic Theory of Gases

139216 An insulated box containing a diatomic gas of molar mass $M$ is moving with a velocity $v$ . The box is suddenly stopped. The resulting change in temperature is (where, $R$ is the gas constant)

1

\frac{M v^{2}}{2 R}

2

\frac{M v^{2}}{3 R}

3

\frac{{Mv}^{2}}{5 R}

4

\frac{2 {Mv}^{2}}{5 R}

Explanation:

C For diatomic gas, $f = 5$
When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises.
The kinetic energy of gas $= \frac{1}{2} {Mv}^{2} \dots$ ...(i)
Change in internal energy, $Δ U = {nC}_{v} Δ T$
For diatomic gas $C_{v} = \frac{5}{2} R$
From equation, (i) and equation (ii), we get -
${nC}_{v} Δ T = n \frac{1}{2} {Mv}^{2}$
$n \frac{5}{2} R Δ T = n \frac{1}{2} {Mv}^{2}$
$5 R Δ T = {Mv}^{2}$
Change in temperature, $Δ T = \frac{{Mv}^{2}}{5 R}$

CG PET -2018

Kinetic Theory of Gases

139217 The rms speed of oxygen molecule at a certain temperature is $600 {ms}^{- 1}$ . If the temperature is doubled and oxygen molecule dissociates into atomic oxygen atoms, the new rms speed is

1

120 {ms}^{- 1}

2

150 {ms}^{- 1}

3

1200 {ms}^{- 1}

4

600 {ms}^{- 1}

[

Explanation:

C Initial Temperature $= T$
Final Temperature $= 2 T$
Velocity $(v_{rms}) = 600 m / \sec$ .
We know,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$600 = \sqrt{\frac{3 RT}{M}}$
Due to dissociation,
$v_{rms} = \sqrt{\frac{3 R (2 T)}{M / 2}}$
On divide equation (i) by equation (ii), we get-
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 RT}{M}}}{\sqrt{\frac{3 R 2 T}{M / 2}}}$
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{12 RT}{M}}}$
$\frac{600}{v_{rms}} = \frac{1}{2} \Rightarrow v_{rms} = 1200 m / s$
The new rms speed is $1200 m / s$ .

AP EAMCET (23.04.2018) Shift-2

Kinetic Theory of Gases

139218 A thermally insulated vessel with nitrogen gas at $27^{\circ} C$ is moving with a velocity of $100 {ms}^{- 1}$ . If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(Assume entire loss in $KE$ of the gas is given as heat to gas and $R = 8.3 {Jmol}^{- 1} K^{- 1}$ )

1 1.1

2 0.93

3 0.5

4 2.25

Explanation:

D Given that,
Temperature, $T = 27^{\circ} C + 273 = 300 K$
$R = 8.3 J / mol K$
Velocity $= 100 m / \sec$ .
There are $n$ mole of $N_{2} = 28 gm$
$= 28 \times 10^{- 3} kg$
$f = 5$
Increment in pressure due to change of temperature is
$Δ PV = nR Δ T$
$Δ P = \frac{nR Δ T}{V}$
From ideal gas equation -
$PV = nRT$
$P = \frac{nRT}{V}$
Equation (i) divided by equation (ii), we get -
$\frac{Δ P}{P} = \frac{nR Δ T}{nRT} [∵ Δ T = \frac{{Mv}^{2}}{fR}]$
$\frac{Δ P}{P} = \frac{{nMv}^{2}}{fnRT}$
$\frac{Δ P}{P} = \frac{{Mv}^{2}}{fRT}$
So, percentage change in pressure is -
$\frac{Δ P}{P} \times 100 = \frac{28 \times 10^{- 3} \times 100 \times 100 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10}{15 \times 8.3}$
$\frac{Δ P}{P} % = \frac{280}{124.5}$
$\frac{Δ P}{P} % = 2.2489$
$\frac{Δ P}{P} % = 2.25 %$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139221 $V_{1}$ is the speed of sound in a diatomic gas at $273^{\circ} C$ and $V_{2}$ is the r.m.s. speed of its molecules at $273 K$ , then $\frac{V_{1}}{V_{2}}$

1

\sqrt{\frac{15}{14}}

2

\sqrt{\frac{14}{15}}

3

\sqrt{\frac{7}{8}}

4

\sqrt{\frac{8}{7}}

[

Explanation:

B Given, $T_{1} = 273^{\circ} C + 273 K = 546 K$ $T_{2} = 273 K$
$γ_{diatomic} = \frac{7}{5}$
We know,
$V_{1} = \sqrt{\frac{γ {RT}_{1}}{M}} and {(V_{2})}_{rms} = \sqrt{\frac{3 {RT}_{2}}{M}}$
Then,
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{\frac{γ {RT}_{1}}{M}}}{\sqrt{\frac{3 {RT}_{2}}{M}}} = \frac{\sqrt{γ T_{1}}}{\sqrt{3 T_{2}}}$
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{7 \times 546}}{\sqrt{5 \times 3 \times 273}} = \sqrt{\frac{14}{15}}$

Kinetic Theory of Gases

139222 The absolute temperature of a gas is increased to 16 times of the original temperature. The rms speed of its molecules will become

1 4 times

2 16 times

3 64 times

4 256 times

Explanation:

A Let ${(V_{rms})}_{1} = V$
$T_{1} = T$
$T_{2} = 16 T$
$\frac{{(V_{rms})}_{1}}{{(V_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{(V)}{{(V_{rms})}_{2}} = \sqrt{\frac{T}{16 T}} = \frac{1}{4}$
${(V_{rms})}_{2} = 4 V$

J and K-CET-2017

Kinetic Theory of Gases

139216 An insulated box containing a diatomic gas of molar mass $M$ is moving with a velocity $v$ . The box is suddenly stopped. The resulting change in temperature is (where, $R$ is the gas constant)

1

\frac{M v^{2}}{2 R}

2

\frac{M v^{2}}{3 R}

3

\frac{{Mv}^{2}}{5 R}

4

\frac{2 {Mv}^{2}}{5 R}

Explanation:

C For diatomic gas, $f = 5$
When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises.
The kinetic energy of gas $= \frac{1}{2} {Mv}^{2} \dots$ ...(i)
Change in internal energy, $Δ U = {nC}_{v} Δ T$
For diatomic gas $C_{v} = \frac{5}{2} R$
From equation, (i) and equation (ii), we get -
${nC}_{v} Δ T = n \frac{1}{2} {Mv}^{2}$
$n \frac{5}{2} R Δ T = n \frac{1}{2} {Mv}^{2}$
$5 R Δ T = {Mv}^{2}$
Change in temperature, $Δ T = \frac{{Mv}^{2}}{5 R}$

CG PET -2018

Kinetic Theory of Gases

139217 The rms speed of oxygen molecule at a certain temperature is $600 {ms}^{- 1}$ . If the temperature is doubled and oxygen molecule dissociates into atomic oxygen atoms, the new rms speed is

1

120 {ms}^{- 1}

2

150 {ms}^{- 1}

3

1200 {ms}^{- 1}

4

600 {ms}^{- 1}

[

Explanation:

C Initial Temperature $= T$
Final Temperature $= 2 T$
Velocity $(v_{rms}) = 600 m / \sec$ .
We know,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$600 = \sqrt{\frac{3 RT}{M}}$
Due to dissociation,
$v_{rms} = \sqrt{\frac{3 R (2 T)}{M / 2}}$
On divide equation (i) by equation (ii), we get-
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 RT}{M}}}{\sqrt{\frac{3 R 2 T}{M / 2}}}$
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{12 RT}{M}}}$
$\frac{600}{v_{rms}} = \frac{1}{2} \Rightarrow v_{rms} = 1200 m / s$
The new rms speed is $1200 m / s$ .

AP EAMCET (23.04.2018) Shift-2

Kinetic Theory of Gases

139218 A thermally insulated vessel with nitrogen gas at $27^{\circ} C$ is moving with a velocity of $100 {ms}^{- 1}$ . If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(Assume entire loss in $KE$ of the gas is given as heat to gas and $R = 8.3 {Jmol}^{- 1} K^{- 1}$ )

1 1.1

2 0.93

3 0.5

4 2.25

Explanation:

D Given that,
Temperature, $T = 27^{\circ} C + 273 = 300 K$
$R = 8.3 J / mol K$
Velocity $= 100 m / \sec$ .
There are $n$ mole of $N_{2} = 28 gm$
$= 28 \times 10^{- 3} kg$
$f = 5$
Increment in pressure due to change of temperature is
$Δ PV = nR Δ T$
$Δ P = \frac{nR Δ T}{V}$
From ideal gas equation -
$PV = nRT$
$P = \frac{nRT}{V}$
Equation (i) divided by equation (ii), we get -
$\frac{Δ P}{P} = \frac{nR Δ T}{nRT} [∵ Δ T = \frac{{Mv}^{2}}{fR}]$
$\frac{Δ P}{P} = \frac{{nMv}^{2}}{fnRT}$
$\frac{Δ P}{P} = \frac{{Mv}^{2}}{fRT}$
So, percentage change in pressure is -
$\frac{Δ P}{P} \times 100 = \frac{28 \times 10^{- 3} \times 100 \times 100 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10}{15 \times 8.3}$
$\frac{Δ P}{P} % = \frac{280}{124.5}$
$\frac{Δ P}{P} % = 2.2489$
$\frac{Δ P}{P} % = 2.25 %$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139221 $V_{1}$ is the speed of sound in a diatomic gas at $273^{\circ} C$ and $V_{2}$ is the r.m.s. speed of its molecules at $273 K$ , then $\frac{V_{1}}{V_{2}}$

1

\sqrt{\frac{15}{14}}

2

\sqrt{\frac{14}{15}}

3

\sqrt{\frac{7}{8}}

4

\sqrt{\frac{8}{7}}

[

Explanation:

B Given, $T_{1} = 273^{\circ} C + 273 K = 546 K$ $T_{2} = 273 K$
$γ_{diatomic} = \frac{7}{5}$
We know,
$V_{1} = \sqrt{\frac{γ {RT}_{1}}{M}} and {(V_{2})}_{rms} = \sqrt{\frac{3 {RT}_{2}}{M}}$
Then,
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{\frac{γ {RT}_{1}}{M}}}{\sqrt{\frac{3 {RT}_{2}}{M}}} = \frac{\sqrt{γ T_{1}}}{\sqrt{3 T_{2}}}$
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{7 \times 546}}{\sqrt{5 \times 3 \times 273}} = \sqrt{\frac{14}{15}}$

Kinetic Theory of Gases

139222 The absolute temperature of a gas is increased to 16 times of the original temperature. The rms speed of its molecules will become

1 4 times

2 16 times

3 64 times

4 256 times

Explanation:

A Let ${(V_{rms})}_{1} = V$
$T_{1} = T$
$T_{2} = 16 T$
$\frac{{(V_{rms})}_{1}}{{(V_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{(V)}{{(V_{rms})}_{2}} = \sqrt{\frac{T}{16 T}} = \frac{1}{4}$
${(V_{rms})}_{2} = 4 V$

J and K-CET-2017

Kinetic Theory of Gases

139216 An insulated box containing a diatomic gas of molar mass $M$ is moving with a velocity $v$ . The box is suddenly stopped. The resulting change in temperature is (where, $R$ is the gas constant)

1

\frac{M v^{2}}{2 R}

2

\frac{M v^{2}}{3 R}

3

\frac{{Mv}^{2}}{5 R}

4

\frac{2 {Mv}^{2}}{5 R}

Explanation:

C For diatomic gas, $f = 5$
When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises.
The kinetic energy of gas $= \frac{1}{2} {Mv}^{2} \dots$ ...(i)
Change in internal energy, $Δ U = {nC}_{v} Δ T$
For diatomic gas $C_{v} = \frac{5}{2} R$
From equation, (i) and equation (ii), we get -
${nC}_{v} Δ T = n \frac{1}{2} {Mv}^{2}$
$n \frac{5}{2} R Δ T = n \frac{1}{2} {Mv}^{2}$
$5 R Δ T = {Mv}^{2}$
Change in temperature, $Δ T = \frac{{Mv}^{2}}{5 R}$

CG PET -2018

Kinetic Theory of Gases

139217 The rms speed of oxygen molecule at a certain temperature is $600 {ms}^{- 1}$ . If the temperature is doubled and oxygen molecule dissociates into atomic oxygen atoms, the new rms speed is

1

120 {ms}^{- 1}

2

150 {ms}^{- 1}

3

1200 {ms}^{- 1}

4

600 {ms}^{- 1}

[

Explanation:

C Initial Temperature $= T$
Final Temperature $= 2 T$
Velocity $(v_{rms}) = 600 m / \sec$ .
We know,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$600 = \sqrt{\frac{3 RT}{M}}$
Due to dissociation,
$v_{rms} = \sqrt{\frac{3 R (2 T)}{M / 2}}$
On divide equation (i) by equation (ii), we get-
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 RT}{M}}}{\sqrt{\frac{3 R 2 T}{M / 2}}}$
$\frac{600}{v_{rms}} = \frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{12 RT}{M}}}$
$\frac{600}{v_{rms}} = \frac{1}{2} \Rightarrow v_{rms} = 1200 m / s$
The new rms speed is $1200 m / s$ .

AP EAMCET (23.04.2018) Shift-2

Kinetic Theory of Gases

139218 A thermally insulated vessel with nitrogen gas at $27^{\circ} C$ is moving with a velocity of $100 {ms}^{- 1}$ . If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(Assume entire loss in $KE$ of the gas is given as heat to gas and $R = 8.3 {Jmol}^{- 1} K^{- 1}$ )

1 1.1

2 0.93

3 0.5

4 2.25

Explanation:

D Given that,
Temperature, $T = 27^{\circ} C + 273 = 300 K$
$R = 8.3 J / mol K$
Velocity $= 100 m / \sec$ .
There are $n$ mole of $N_{2} = 28 gm$
$= 28 \times 10^{- 3} kg$
$f = 5$
Increment in pressure due to change of temperature is
$Δ PV = nR Δ T$
$Δ P = \frac{nR Δ T}{V}$
From ideal gas equation -
$PV = nRT$
$P = \frac{nRT}{V}$
Equation (i) divided by equation (ii), we get -
$\frac{Δ P}{P} = \frac{nR Δ T}{nRT} [∵ Δ T = \frac{{Mv}^{2}}{fR}]$
$\frac{Δ P}{P} = \frac{{nMv}^{2}}{fnRT}$
$\frac{Δ P}{P} = \frac{{Mv}^{2}}{fRT}$
So, percentage change in pressure is -
$\frac{Δ P}{P} \times 100 = \frac{28 \times 10^{- 3} \times 100 \times 100 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10 \times 100}{5 \times 8.3 \times 300}$
$\frac{Δ P}{P} % = \frac{28 \times 10}{15 \times 8.3}$
$\frac{Δ P}{P} % = \frac{280}{124.5}$
$\frac{Δ P}{P} % = 2.2489$
$\frac{Δ P}{P} % = 2.25 %$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139221 $V_{1}$ is the speed of sound in a diatomic gas at $273^{\circ} C$ and $V_{2}$ is the r.m.s. speed of its molecules at $273 K$ , then $\frac{V_{1}}{V_{2}}$

1

\sqrt{\frac{15}{14}}

2

\sqrt{\frac{14}{15}}

3

\sqrt{\frac{7}{8}}

4

\sqrt{\frac{8}{7}}

[

Explanation:

B Given, $T_{1} = 273^{\circ} C + 273 K = 546 K$ $T_{2} = 273 K$
$γ_{diatomic} = \frac{7}{5}$
We know,
$V_{1} = \sqrt{\frac{γ {RT}_{1}}{M}} and {(V_{2})}_{rms} = \sqrt{\frac{3 {RT}_{2}}{M}}$
Then,
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{\frac{γ {RT}_{1}}{M}}}{\sqrt{\frac{3 {RT}_{2}}{M}}} = \frac{\sqrt{γ T_{1}}}{\sqrt{3 T_{2}}}$
$\frac{V_{1}}{V_{2}} = \frac{\sqrt{7 \times 546}}{\sqrt{5 \times 3 \times 273}} = \sqrt{\frac{14}{15}}$

Kinetic Theory of Gases

139222 The absolute temperature of a gas is increased to 16 times of the original temperature. The rms speed of its molecules will become

1 4 times

2 16 times

3 64 times

4 256 times

Explanation:

A Let ${(V_{rms})}_{1} = V$
$T_{1} = T$
$T_{2} = 16 T$
$\frac{{(V_{rms})}_{1}}{{(V_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{(V)}{{(V_{rms})}_{2}} = \sqrt{\frac{T}{16 T}} = \frac{1}{4}$
${(V_{rms})}_{2} = 4 V$

J and K-CET-2017