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Kinetic Theory of Gases

139151 An ideal gas at pressure $P_{0}$ undergoes an isothermal expansion until its volume is 8.0 times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $γ$ $= 4 / 3$ , then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1 1.44

2 1.68

3 2.0

4 1.2

Explanation:

C
Given,
Initial volume $(V_{0}) = V$
Final volume $(V_{1}) = 8 V$
For isothermal process,
$P_{0} V_{0} = P_{1} V_{1} \Rightarrow P_{0} V = P_{1} \times 8 V$
$P_{1} = \frac{P_{0}}{8}$
For adiabatic process,
Initial volume $(V_{1}) = 8 V$ , Final volume $(V_{2}) = V$
Initial pressure $(P_{1}) = \frac{P_{0}}{8}, γ = \frac{4}{3}$
$P_{1} {(V_{1})}^{γ} = P_{2} {(V_{2})}^{γ}$
$P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = \frac{P_{0}}{8} \times {(\frac{8 V}{V})}^{4 / 3}$
$P_{2} = \frac{P_{0}}{8} \times 16 = 2 P_{0}$
Then, $\frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{γ}{γ - 1}}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$= {(\frac{2 P_{0}}{P_{0} / 8})}^{\frac{\frac{4}{3} - 1}{4 / 3}}$
$= (16)^{1 / 4}$
$\frac{T_{2}}{T_{1}} = 2$
$∵$ Average kinetic energy $(v_{avg}) = \sqrt{\frac{3 R T}{M}}$
$v_{avg}^{2} \propto T$
Final state of average kinetic energy,
${(K_{2})}_{avg} = \frac{1}{2} m {(v_{2})}_{rms}^{2}$
Initial state of average kinetic energy,
${(K_{1})}_{avg} = \frac{1}{2} m {(v_{1})}_{rms}^{2}$
Ratio, $\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = {(\frac{v_{2}}{v_{1}})}_{rms}^{2} (∵ v_{avg}^{2} \propto T)$
$\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = \frac{T_{2}}{T_{1}} = 2$

Kinetic Theory of Gases

139152 At what temperature is the root mean square (rms) speed of Neon gas atoms is equal to the rms speed of Helium gas atom at $- 33^{\circ} C$ ?
(atomic mass of $Ne = 20.2 u$ . and that of $He =$ $4.0 u)$
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1

1208 K

2

1210 K

3

1212 K

4

1220 K

Explanation:

C ${(v_{rms})}_{Ne} = {(v_{rms})}_{He}$
Atomic mass of $Ne (M_{Ne}) = 20.2 u$
Atomic mass of $He (M_{He}) = 4.0 u$
$T_{He} = 273 - 33 = 240 K$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{Ne})}_{rms}}{(v_{He})} = \frac{\sqrt{\frac{3 {RT}_{Ne}}{M_{Ne}}}}{\sqrt{\frac{3 {RT}_{He}}{M_{He}}}}$
$1 = \sqrt{\frac{M_{He} \times T_{Ne}}{M_{Ne} \times T_{He}}}$
$1 = \frac{4 \times T_{Ne}}{20.2 \times 240}$
$T_{Ne} = 1212 K$

Kinetic Theory of Gases

139153 Two gases $A$ and $B$ are having molar masses of $M_{1}$ and $M_{2}$ and temperature $T_{1}$ and $T_{2}$ respectively. The rms speed of gas $A$ is twice the rms speed of gas $B$ . If the ratio of $M_{1}$ to $M_{2}$ is $2 : 1$ , then the ratio of $T_{1}$ to $T_{2}$ is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 30.07.2022,Shift-II#

1

8 : 1

2

1 : 8

3

4 : 1

4

1 : 4

Explanation:

A Given,
${(v_{1})}_{mms} = 2 {(v_{2})}_{mms} \Rightarrow \frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{2}{1}$
$\frac{M_{1}}{M_{2}} = \frac{2}{1}$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{\sqrt{\frac{3 {RT}_{1}}{M_{1}}}}{\sqrt{\frac{3 {RT}_{2}}{M_{2}}}} = \sqrt{\frac{M_{2} T_{1}}{M_{1} T_{2}}}$
$\frac{2 {(v_{2})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{2 T_{2}}}$
$4 = \frac{T_{1}}{2 T_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{8}{1} \Rightarrow T_{1} : T_{2} = 8 : 1$

Kinetic Theory of Gases

139154 A gaseous mixture consists of molecules of type $A, B$ and $C$ with masses $M_{A} > M_{B} > M_{C}$ . The correct relation between their average kinetic energy is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 31.07.2022,Shift-II#

1

K_{A} > K_{B} > K_{C}

2

K_{A} < K_{B} < K_{C}

3

K_{A} = K_{B} = K_{C}

4

K_{A} = K_{B} < K_{C}

Explanation:

C Given,
$M_{A} > M_{B} > M_{C}$
$KE = \frac{3}{2} KT \Rightarrow K \propto \sqrt{T}$
Average KE depends on only temperature. Since the mixture will be at same temperature $(T)$ .
So average kinetic energy will be same for all types of molecules.
$K_{A} = K_{B} = K_{C}$

Kinetic Theory of Gases

139151 An ideal gas at pressure $P_{0}$ undergoes an isothermal expansion until its volume is 8.0 times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $γ$ $= 4 / 3$ , then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1 1.44

2 1.68

3 2.0

4 1.2

Explanation:

C
Given,
Initial volume $(V_{0}) = V$
Final volume $(V_{1}) = 8 V$
For isothermal process,
$P_{0} V_{0} = P_{1} V_{1} \Rightarrow P_{0} V = P_{1} \times 8 V$
$P_{1} = \frac{P_{0}}{8}$
For adiabatic process,
Initial volume $(V_{1}) = 8 V$ , Final volume $(V_{2}) = V$
Initial pressure $(P_{1}) = \frac{P_{0}}{8}, γ = \frac{4}{3}$
$P_{1} {(V_{1})}^{γ} = P_{2} {(V_{2})}^{γ}$
$P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = \frac{P_{0}}{8} \times {(\frac{8 V}{V})}^{4 / 3}$
$P_{2} = \frac{P_{0}}{8} \times 16 = 2 P_{0}$
Then, $\frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{γ}{γ - 1}}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$= {(\frac{2 P_{0}}{P_{0} / 8})}^{\frac{\frac{4}{3} - 1}{4 / 3}}$
$= (16)^{1 / 4}$
$\frac{T_{2}}{T_{1}} = 2$
$∵$ Average kinetic energy $(v_{avg}) = \sqrt{\frac{3 R T}{M}}$
$v_{avg}^{2} \propto T$
Final state of average kinetic energy,
${(K_{2})}_{avg} = \frac{1}{2} m {(v_{2})}_{rms}^{2}$
Initial state of average kinetic energy,
${(K_{1})}_{avg} = \frac{1}{2} m {(v_{1})}_{rms}^{2}$
Ratio, $\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = {(\frac{v_{2}}{v_{1}})}_{rms}^{2} (∵ v_{avg}^{2} \propto T)$
$\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = \frac{T_{2}}{T_{1}} = 2$

Kinetic Theory of Gases

139152 At what temperature is the root mean square (rms) speed of Neon gas atoms is equal to the rms speed of Helium gas atom at $- 33^{\circ} C$ ?
(atomic mass of $Ne = 20.2 u$ . and that of $He =$ $4.0 u)$
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1

1208 K

2

1210 K

3

1212 K

4

1220 K

Explanation:

C ${(v_{rms})}_{Ne} = {(v_{rms})}_{He}$
Atomic mass of $Ne (M_{Ne}) = 20.2 u$
Atomic mass of $He (M_{He}) = 4.0 u$
$T_{He} = 273 - 33 = 240 K$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{Ne})}_{rms}}{(v_{He})} = \frac{\sqrt{\frac{3 {RT}_{Ne}}{M_{Ne}}}}{\sqrt{\frac{3 {RT}_{He}}{M_{He}}}}$
$1 = \sqrt{\frac{M_{He} \times T_{Ne}}{M_{Ne} \times T_{He}}}$
$1 = \frac{4 \times T_{Ne}}{20.2 \times 240}$
$T_{Ne} = 1212 K$

Kinetic Theory of Gases

139153 Two gases $A$ and $B$ are having molar masses of $M_{1}$ and $M_{2}$ and temperature $T_{1}$ and $T_{2}$ respectively. The rms speed of gas $A$ is twice the rms speed of gas $B$ . If the ratio of $M_{1}$ to $M_{2}$ is $2 : 1$ , then the ratio of $T_{1}$ to $T_{2}$ is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 30.07.2022,Shift-II#

1

8 : 1

2

1 : 8

3

4 : 1

4

1 : 4

Explanation:

A Given,
${(v_{1})}_{mms} = 2 {(v_{2})}_{mms} \Rightarrow \frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{2}{1}$
$\frac{M_{1}}{M_{2}} = \frac{2}{1}$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{\sqrt{\frac{3 {RT}_{1}}{M_{1}}}}{\sqrt{\frac{3 {RT}_{2}}{M_{2}}}} = \sqrt{\frac{M_{2} T_{1}}{M_{1} T_{2}}}$
$\frac{2 {(v_{2})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{2 T_{2}}}$
$4 = \frac{T_{1}}{2 T_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{8}{1} \Rightarrow T_{1} : T_{2} = 8 : 1$

Kinetic Theory of Gases

139154 A gaseous mixture consists of molecules of type $A, B$ and $C$ with masses $M_{A} > M_{B} > M_{C}$ . The correct relation between their average kinetic energy is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 31.07.2022,Shift-II#

1

K_{A} > K_{B} > K_{C}

2

K_{A} < K_{B} < K_{C}

3

K_{A} = K_{B} = K_{C}

4

K_{A} = K_{B} < K_{C}

Explanation:

C Given,
$M_{A} > M_{B} > M_{C}$
$KE = \frac{3}{2} KT \Rightarrow K \propto \sqrt{T}$
Average KE depends on only temperature. Since the mixture will be at same temperature $(T)$ .
So average kinetic energy will be same for all types of molecules.
$K_{A} = K_{B} = K_{C}$

Kinetic Theory of Gases

139151 An ideal gas at pressure $P_{0}$ undergoes an isothermal expansion until its volume is 8.0 times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $γ$ $= 4 / 3$ , then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1 1.44

2 1.68

3 2.0

4 1.2

Explanation:

C
Given,
Initial volume $(V_{0}) = V$
Final volume $(V_{1}) = 8 V$
For isothermal process,
$P_{0} V_{0} = P_{1} V_{1} \Rightarrow P_{0} V = P_{1} \times 8 V$
$P_{1} = \frac{P_{0}}{8}$
For adiabatic process,
Initial volume $(V_{1}) = 8 V$ , Final volume $(V_{2}) = V$
Initial pressure $(P_{1}) = \frac{P_{0}}{8}, γ = \frac{4}{3}$
$P_{1} {(V_{1})}^{γ} = P_{2} {(V_{2})}^{γ}$
$P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = \frac{P_{0}}{8} \times {(\frac{8 V}{V})}^{4 / 3}$
$P_{2} = \frac{P_{0}}{8} \times 16 = 2 P_{0}$
Then, $\frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{γ}{γ - 1}}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$= {(\frac{2 P_{0}}{P_{0} / 8})}^{\frac{\frac{4}{3} - 1}{4 / 3}}$
$= (16)^{1 / 4}$
$\frac{T_{2}}{T_{1}} = 2$
$∵$ Average kinetic energy $(v_{avg}) = \sqrt{\frac{3 R T}{M}}$
$v_{avg}^{2} \propto T$
Final state of average kinetic energy,
${(K_{2})}_{avg} = \frac{1}{2} m {(v_{2})}_{rms}^{2}$
Initial state of average kinetic energy,
${(K_{1})}_{avg} = \frac{1}{2} m {(v_{1})}_{rms}^{2}$
Ratio, $\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = {(\frac{v_{2}}{v_{1}})}_{rms}^{2} (∵ v_{avg}^{2} \propto T)$
$\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = \frac{T_{2}}{T_{1}} = 2$

Kinetic Theory of Gases

139152 At what temperature is the root mean square (rms) speed of Neon gas atoms is equal to the rms speed of Helium gas atom at $- 33^{\circ} C$ ?
(atomic mass of $Ne = 20.2 u$ . and that of $He =$ $4.0 u)$
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1

1208 K

2

1210 K

3

1212 K

4

1220 K

Explanation:

C ${(v_{rms})}_{Ne} = {(v_{rms})}_{He}$
Atomic mass of $Ne (M_{Ne}) = 20.2 u$
Atomic mass of $He (M_{He}) = 4.0 u$
$T_{He} = 273 - 33 = 240 K$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{Ne})}_{rms}}{(v_{He})} = \frac{\sqrt{\frac{3 {RT}_{Ne}}{M_{Ne}}}}{\sqrt{\frac{3 {RT}_{He}}{M_{He}}}}$
$1 = \sqrt{\frac{M_{He} \times T_{Ne}}{M_{Ne} \times T_{He}}}$
$1 = \frac{4 \times T_{Ne}}{20.2 \times 240}$
$T_{Ne} = 1212 K$

Kinetic Theory of Gases

139153 Two gases $A$ and $B$ are having molar masses of $M_{1}$ and $M_{2}$ and temperature $T_{1}$ and $T_{2}$ respectively. The rms speed of gas $A$ is twice the rms speed of gas $B$ . If the ratio of $M_{1}$ to $M_{2}$ is $2 : 1$ , then the ratio of $T_{1}$ to $T_{2}$ is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 30.07.2022,Shift-II#

1

8 : 1

2

1 : 8

3

4 : 1

4

1 : 4

Explanation:

A Given,
${(v_{1})}_{mms} = 2 {(v_{2})}_{mms} \Rightarrow \frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{2}{1}$
$\frac{M_{1}}{M_{2}} = \frac{2}{1}$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{\sqrt{\frac{3 {RT}_{1}}{M_{1}}}}{\sqrt{\frac{3 {RT}_{2}}{M_{2}}}} = \sqrt{\frac{M_{2} T_{1}}{M_{1} T_{2}}}$
$\frac{2 {(v_{2})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{2 T_{2}}}$
$4 = \frac{T_{1}}{2 T_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{8}{1} \Rightarrow T_{1} : T_{2} = 8 : 1$

Kinetic Theory of Gases

139154 A gaseous mixture consists of molecules of type $A, B$ and $C$ with masses $M_{A} > M_{B} > M_{C}$ . The correct relation between their average kinetic energy is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 31.07.2022,Shift-II#

1

K_{A} > K_{B} > K_{C}

2

K_{A} < K_{B} < K_{C}

3

K_{A} = K_{B} = K_{C}

4

K_{A} = K_{B} < K_{C}

Explanation:

C Given,
$M_{A} > M_{B} > M_{C}$
$KE = \frac{3}{2} KT \Rightarrow K \propto \sqrt{T}$
Average KE depends on only temperature. Since the mixture will be at same temperature $(T)$ .
So average kinetic energy will be same for all types of molecules.
$K_{A} = K_{B} = K_{C}$

Kinetic Theory of Gases

139151 An ideal gas at pressure $P_{0}$ undergoes an isothermal expansion until its volume is 8.0 times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $γ$ $= 4 / 3$ , then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1 1.44

2 1.68

3 2.0

4 1.2

Explanation:

C
Given,
Initial volume $(V_{0}) = V$
Final volume $(V_{1}) = 8 V$
For isothermal process,
$P_{0} V_{0} = P_{1} V_{1} \Rightarrow P_{0} V = P_{1} \times 8 V$
$P_{1} = \frac{P_{0}}{8}$
For adiabatic process,
Initial volume $(V_{1}) = 8 V$ , Final volume $(V_{2}) = V$
Initial pressure $(P_{1}) = \frac{P_{0}}{8}, γ = \frac{4}{3}$
$P_{1} {(V_{1})}^{γ} = P_{2} {(V_{2})}^{γ}$
$P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = \frac{P_{0}}{8} \times {(\frac{8 V}{V})}^{4 / 3}$
$P_{2} = \frac{P_{0}}{8} \times 16 = 2 P_{0}$
Then, $\frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{γ}{γ - 1}}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$= {(\frac{2 P_{0}}{P_{0} / 8})}^{\frac{\frac{4}{3} - 1}{4 / 3}}$
$= (16)^{1 / 4}$
$\frac{T_{2}}{T_{1}} = 2$
$∵$ Average kinetic energy $(v_{avg}) = \sqrt{\frac{3 R T}{M}}$
$v_{avg}^{2} \propto T$
Final state of average kinetic energy,
${(K_{2})}_{avg} = \frac{1}{2} m {(v_{2})}_{rms}^{2}$
Initial state of average kinetic energy,
${(K_{1})}_{avg} = \frac{1}{2} m {(v_{1})}_{rms}^{2}$
Ratio, $\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = {(\frac{v_{2}}{v_{1}})}_{rms}^{2} (∵ v_{avg}^{2} \propto T)$
$\frac{{(K_{2})}_{avg}}{{(K_{1})}_{avg}} = \frac{T_{2}}{T_{1}} = 2$

Kinetic Theory of Gases

139152 At what temperature is the root mean square (rms) speed of Neon gas atoms is equal to the rms speed of Helium gas atom at $- 33^{\circ} C$ ?
(atomic mass of $Ne = 20.2 u$ . and that of $He =$ $4.0 u)$
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 19.07.2022,Shift-II#

1

1208 K

2

1210 K

3

1212 K

4

1220 K

Explanation:

C ${(v_{rms})}_{Ne} = {(v_{rms})}_{He}$
Atomic mass of $Ne (M_{Ne}) = 20.2 u$
Atomic mass of $He (M_{He}) = 4.0 u$
$T_{He} = 273 - 33 = 240 K$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{Ne})}_{rms}}{(v_{He})} = \frac{\sqrt{\frac{3 {RT}_{Ne}}{M_{Ne}}}}{\sqrt{\frac{3 {RT}_{He}}{M_{He}}}}$
$1 = \sqrt{\frac{M_{He} \times T_{Ne}}{M_{Ne} \times T_{He}}}$
$1 = \frac{4 \times T_{Ne}}{20.2 \times 240}$
$T_{Ne} = 1212 K$

Kinetic Theory of Gases

139153 Two gases $A$ and $B$ are having molar masses of $M_{1}$ and $M_{2}$ and temperature $T_{1}$ and $T_{2}$ respectively. The rms speed of gas $A$ is twice the rms speed of gas $B$ . If the ratio of $M_{1}$ to $M_{2}$ is $2 : 1$ , then the ratio of $T_{1}$ to $T_{2}$ is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 30.07.2022,Shift-II#

1

8 : 1

2

1 : 8

3

4 : 1

4

1 : 4

Explanation:

A Given,
${(v_{1})}_{mms} = 2 {(v_{2})}_{mms} \Rightarrow \frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{2}{1}$
$\frac{M_{1}}{M_{2}} = \frac{2}{1}$
$∵ v_{rms} = \sqrt{\frac{3 RT}{M}}$
$\frac{{(v_{1})}_{mms}}{{(v_{2})}_{rms}} = \frac{\sqrt{\frac{3 {RT}_{1}}{M_{1}}}}{\sqrt{\frac{3 {RT}_{2}}{M_{2}}}} = \sqrt{\frac{M_{2} T_{1}}{M_{1} T_{2}}}$
$\frac{2 {(v_{2})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{2 T_{2}}}$
$4 = \frac{T_{1}}{2 T_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{8}{1} \Rightarrow T_{1} : T_{2} = 8 : 1$

Kinetic Theory of Gases

139154 A gaseous mixture consists of molecules of type $A, B$ and $C$ with masses $M_{A} > M_{B} > M_{C}$ . The correct relation between their average kinetic energy is
#[Qdiff: Hard, QCat: Numerical Based, examname: TS EAMCET 31.07.2022,Shift-II#

1

K_{A} > K_{B} > K_{C}

2

K_{A} < K_{B} < K_{C}

3

K_{A} = K_{B} = K_{C}

4

K_{A} = K_{B} < K_{C}

Explanation:

C Given,
$M_{A} > M_{B} > M_{C}$
$KE = \frac{3}{2} KT \Rightarrow K \propto \sqrt{T}$
Average KE depends on only temperature. Since the mixture will be at same temperature $(T)$ .
So average kinetic energy will be same for all types of molecules.
$K_{A} = K_{B} = K_{C}$

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