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Kinetic Theory of Gases

139118 The temperature, at which the rms velocity of hydrogen is four times of its value at NTP is

1

819^{\circ} C

2

1092^{\circ} C

3

4368^{\circ} C

4

4095^{\circ} C

Explanation:

D We know that,
$T_{1} = 273 K$
r.m.s. velocity of hydrogen, $v_{r.m.s.} = v$
$v_{r.m.s.} = \sqrt{\frac{3 RT}{M}}$
Here R, M are constant
So,
$\frac{{(v_{r.m.s.})}_{1}}{v_{r.m.s.}} \propto \sqrt{T}$
$\frac{{(v_{r.m.s.})}_{2}}{\frac{T_{1}}{T_{2}}}$
$\frac{v}{4 v} = \sqrt{\frac{273}{T_{2}}}$
$\frac{1}{16} = \frac{273}{T_{2}}$
$T_{2} = 4368 K = (4368 - 273)^{\circ} C$
$T_{2} = 4095^{\circ} C$

MHT-CET 2005

Kinetic Theory of Gases

139119 The mean kinetic energy of monoatomic gas molecules under standard conditions is $⟨ E_{1} ⟩$ . If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to $⟨ E_{2} ⟩$ . The ratio $\frac{⟨ E_{2} ⟩}{⟨ E_{1} ⟩}$ is

1 2

2 4

3 6

4 8

Explanation:

B Given, $V_{2} = \frac{1}{8} V_{1}$
For monoatomic gas molecules
$γ = \frac{5}{3}$
Mean kinetic energy per molecule
$E = \frac{3}{2} K_{B} T$
$∴ E \propto T$
Here
$T_{1} = T, V_{2} = \frac{V_{1}}{8}$
$∴ \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} (E in terms of T)$
$\frac{E_{2}}{E_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V_{1}}{\frac{1}{8} V_{1}})}^{\frac{5}{3} - 1} = (8)^{2 / 3}$
$\frac{E_{2}}{E_{1}} = 4$

Shift-2]

Kinetic Theory of Gases

139120 The temperature at which the r.m.s speed of molecules in hydrogen gas will be double of its initial value at $27^{\circ} C$ is

1

300^{\circ} C

2

1473^{\circ} C

3

927^{\circ} C

4

546^{\circ} C

Explanation:

C Given,
Initial temperature $= 27^{\circ} C = 300 K$
Initial velocity $= V$
Final velocity $= 2 V$
We know that,
$V_{rms} = \sqrt{\frac{3 R RT}{M}}$
$V \propto T^{1 / 2}$
$\frac{V_{1}}{T_{1}^{1 / 2}} = \frac{V_{2}}{T_{2}^{1 / 2}}$
$\frac{V}{(300)^{1 / 2}} = \frac{2 V}{{(T_{2})}^{1 / 2}}$
${(\frac{T_{2}}{300})}^{1 / 2} = 2$
$\frac{T_{2}}{300} = 4$
$T_{2} = 1200 K$
$T_{2} = 1200 - 273$
$T_{2} = 927^{\circ} C$
$∴ T_{2} = 1200 - 273$

Shift-II

Kinetic Theory of Gases

139121 The r.m.s. velocity of hydrogen molecules at temperature $T$ is seven times the r.m.s. velocity of nitrogen molecules at $300 K$ . This temperature $T$ is (Molecular weights of hydrogen and nitrogen are 2 and 28 respectively)

1

1350 K

2

1700 K

3

1050 K

4

2100 K

Explanation:

C We know that,
$V_{rms} = \sqrt{\frac{3 RT}{M}}$
${(V_{rms})}_{H_{2}} = 7 \times {(V_{rms})}_{N_{2}}$
$\sqrt{\frac{3 {RT}_{H_{2}}}{M_{H_{2}}}} = 7 \times \sqrt{\frac{3 {RT}_{N_{2}}}{M_{N_{2}}}}$
$∴ \frac{T_{H_{2}}}{2} = 49 \times \frac{T_{N_{2}}}{28}$
$T_{H_{2}} = 7 \times \frac{T_{N_{2}}}{2}$
$= 7 \times \frac{300}{2}$
$= 7 \times 150$
$= 1050 K$

MHT-CET 2020

Kinetic Theory of Gases

139118 The temperature, at which the rms velocity of hydrogen is four times of its value at NTP is

1

819^{\circ} C

2

1092^{\circ} C

3

4368^{\circ} C

4

4095^{\circ} C

Explanation:

D We know that,
$T_{1} = 273 K$
r.m.s. velocity of hydrogen, $v_{r.m.s.} = v$
$v_{r.m.s.} = \sqrt{\frac{3 RT}{M}}$
Here R, M are constant
So,
$\frac{{(v_{r.m.s.})}_{1}}{v_{r.m.s.}} \propto \sqrt{T}$
$\frac{{(v_{r.m.s.})}_{2}}{\frac{T_{1}}{T_{2}}}$
$\frac{v}{4 v} = \sqrt{\frac{273}{T_{2}}}$
$\frac{1}{16} = \frac{273}{T_{2}}$
$T_{2} = 4368 K = (4368 - 273)^{\circ} C$
$T_{2} = 4095^{\circ} C$

MHT-CET 2005

Kinetic Theory of Gases

139119 The mean kinetic energy of monoatomic gas molecules under standard conditions is $⟨ E_{1} ⟩$ . If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to $⟨ E_{2} ⟩$ . The ratio $\frac{⟨ E_{2} ⟩}{⟨ E_{1} ⟩}$ is

1 2

2 4

3 6

4 8

Explanation:

B Given, $V_{2} = \frac{1}{8} V_{1}$
For monoatomic gas molecules
$γ = \frac{5}{3}$
Mean kinetic energy per molecule
$E = \frac{3}{2} K_{B} T$
$∴ E \propto T$
Here
$T_{1} = T, V_{2} = \frac{V_{1}}{8}$
$∴ \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} (E in terms of T)$
$\frac{E_{2}}{E_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V_{1}}{\frac{1}{8} V_{1}})}^{\frac{5}{3} - 1} = (8)^{2 / 3}$
$\frac{E_{2}}{E_{1}} = 4$

Shift-2]

Kinetic Theory of Gases

139120 The temperature at which the r.m.s speed of molecules in hydrogen gas will be double of its initial value at $27^{\circ} C$ is

1

300^{\circ} C

2

1473^{\circ} C

3

927^{\circ} C

4

546^{\circ} C

Explanation:

C Given,
Initial temperature $= 27^{\circ} C = 300 K$
Initial velocity $= V$
Final velocity $= 2 V$
We know that,
$V_{rms} = \sqrt{\frac{3 R RT}{M}}$
$V \propto T^{1 / 2}$
$\frac{V_{1}}{T_{1}^{1 / 2}} = \frac{V_{2}}{T_{2}^{1 / 2}}$
$\frac{V}{(300)^{1 / 2}} = \frac{2 V}{{(T_{2})}^{1 / 2}}$
${(\frac{T_{2}}{300})}^{1 / 2} = 2$
$\frac{T_{2}}{300} = 4$
$T_{2} = 1200 K$
$T_{2} = 1200 - 273$
$T_{2} = 927^{\circ} C$
$∴ T_{2} = 1200 - 273$

Shift-II

Kinetic Theory of Gases

139121 The r.m.s. velocity of hydrogen molecules at temperature $T$ is seven times the r.m.s. velocity of nitrogen molecules at $300 K$ . This temperature $T$ is (Molecular weights of hydrogen and nitrogen are 2 and 28 respectively)

1

1350 K

2

1700 K

3

1050 K

4

2100 K

Explanation:

C We know that,
$V_{rms} = \sqrt{\frac{3 RT}{M}}$
${(V_{rms})}_{H_{2}} = 7 \times {(V_{rms})}_{N_{2}}$
$\sqrt{\frac{3 {RT}_{H_{2}}}{M_{H_{2}}}} = 7 \times \sqrt{\frac{3 {RT}_{N_{2}}}{M_{N_{2}}}}$
$∴ \frac{T_{H_{2}}}{2} = 49 \times \frac{T_{N_{2}}}{28}$
$T_{H_{2}} = 7 \times \frac{T_{N_{2}}}{2}$
$= 7 \times \frac{300}{2}$
$= 7 \times 150$
$= 1050 K$

MHT-CET 2020

Kinetic Theory of Gases

139118 The temperature, at which the rms velocity of hydrogen is four times of its value at NTP is

1

819^{\circ} C

2

1092^{\circ} C

3

4368^{\circ} C

4

4095^{\circ} C

Explanation:

D We know that,
$T_{1} = 273 K$
r.m.s. velocity of hydrogen, $v_{r.m.s.} = v$
$v_{r.m.s.} = \sqrt{\frac{3 RT}{M}}$
Here R, M are constant
So,
$\frac{{(v_{r.m.s.})}_{1}}{v_{r.m.s.}} \propto \sqrt{T}$
$\frac{{(v_{r.m.s.})}_{2}}{\frac{T_{1}}{T_{2}}}$
$\frac{v}{4 v} = \sqrt{\frac{273}{T_{2}}}$
$\frac{1}{16} = \frac{273}{T_{2}}$
$T_{2} = 4368 K = (4368 - 273)^{\circ} C$
$T_{2} = 4095^{\circ} C$

MHT-CET 2005

Kinetic Theory of Gases

139119 The mean kinetic energy of monoatomic gas molecules under standard conditions is $⟨ E_{1} ⟩$ . If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to $⟨ E_{2} ⟩$ . The ratio $\frac{⟨ E_{2} ⟩}{⟨ E_{1} ⟩}$ is

1 2

2 4

3 6

4 8

Explanation:

B Given, $V_{2} = \frac{1}{8} V_{1}$
For monoatomic gas molecules
$γ = \frac{5}{3}$
Mean kinetic energy per molecule
$E = \frac{3}{2} K_{B} T$
$∴ E \propto T$
Here
$T_{1} = T, V_{2} = \frac{V_{1}}{8}$
$∴ \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} (E in terms of T)$
$\frac{E_{2}}{E_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V_{1}}{\frac{1}{8} V_{1}})}^{\frac{5}{3} - 1} = (8)^{2 / 3}$
$\frac{E_{2}}{E_{1}} = 4$

Shift-2]

Kinetic Theory of Gases

139120 The temperature at which the r.m.s speed of molecules in hydrogen gas will be double of its initial value at $27^{\circ} C$ is

1

300^{\circ} C

2

1473^{\circ} C

3

927^{\circ} C

4

546^{\circ} C

Explanation:

C Given,
Initial temperature $= 27^{\circ} C = 300 K$
Initial velocity $= V$
Final velocity $= 2 V$
We know that,
$V_{rms} = \sqrt{\frac{3 R RT}{M}}$
$V \propto T^{1 / 2}$
$\frac{V_{1}}{T_{1}^{1 / 2}} = \frac{V_{2}}{T_{2}^{1 / 2}}$
$\frac{V}{(300)^{1 / 2}} = \frac{2 V}{{(T_{2})}^{1 / 2}}$
${(\frac{T_{2}}{300})}^{1 / 2} = 2$
$\frac{T_{2}}{300} = 4$
$T_{2} = 1200 K$
$T_{2} = 1200 - 273$
$T_{2} = 927^{\circ} C$
$∴ T_{2} = 1200 - 273$

Shift-II

Kinetic Theory of Gases

139121 The r.m.s. velocity of hydrogen molecules at temperature $T$ is seven times the r.m.s. velocity of nitrogen molecules at $300 K$ . This temperature $T$ is (Molecular weights of hydrogen and nitrogen are 2 and 28 respectively)

1

1350 K

2

1700 K

3

1050 K

4

2100 K

Explanation:

C We know that,
$V_{rms} = \sqrt{\frac{3 RT}{M}}$
${(V_{rms})}_{H_{2}} = 7 \times {(V_{rms})}_{N_{2}}$
$\sqrt{\frac{3 {RT}_{H_{2}}}{M_{H_{2}}}} = 7 \times \sqrt{\frac{3 {RT}_{N_{2}}}{M_{N_{2}}}}$
$∴ \frac{T_{H_{2}}}{2} = 49 \times \frac{T_{N_{2}}}{28}$
$T_{H_{2}} = 7 \times \frac{T_{N_{2}}}{2}$
$= 7 \times \frac{300}{2}$
$= 7 \times 150$
$= 1050 K$

MHT-CET 2020

Kinetic Theory of Gases

139118 The temperature, at which the rms velocity of hydrogen is four times of its value at NTP is

1

819^{\circ} C

2

1092^{\circ} C

3

4368^{\circ} C

4

4095^{\circ} C

Explanation:

D We know that,
$T_{1} = 273 K$
r.m.s. velocity of hydrogen, $v_{r.m.s.} = v$
$v_{r.m.s.} = \sqrt{\frac{3 RT}{M}}$
Here R, M are constant
So,
$\frac{{(v_{r.m.s.})}_{1}}{v_{r.m.s.}} \propto \sqrt{T}$
$\frac{{(v_{r.m.s.})}_{2}}{\frac{T_{1}}{T_{2}}}$
$\frac{v}{4 v} = \sqrt{\frac{273}{T_{2}}}$
$\frac{1}{16} = \frac{273}{T_{2}}$
$T_{2} = 4368 K = (4368 - 273)^{\circ} C$
$T_{2} = 4095^{\circ} C$

MHT-CET 2005

Kinetic Theory of Gases

139119 The mean kinetic energy of monoatomic gas molecules under standard conditions is $⟨ E_{1} ⟩$ . If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to $⟨ E_{2} ⟩$ . The ratio $\frac{⟨ E_{2} ⟩}{⟨ E_{1} ⟩}$ is

1 2

2 4

3 6

4 8

Explanation:

B Given, $V_{2} = \frac{1}{8} V_{1}$
For monoatomic gas molecules
$γ = \frac{5}{3}$
Mean kinetic energy per molecule
$E = \frac{3}{2} K_{B} T$
$∴ E \propto T$
Here
$T_{1} = T, V_{2} = \frac{V_{1}}{8}$
$∴ \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} (E in terms of T)$
$\frac{E_{2}}{E_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V_{1}}{\frac{1}{8} V_{1}})}^{\frac{5}{3} - 1} = (8)^{2 / 3}$
$\frac{E_{2}}{E_{1}} = 4$

Shift-2]

Kinetic Theory of Gases

139120 The temperature at which the r.m.s speed of molecules in hydrogen gas will be double of its initial value at $27^{\circ} C$ is

1

300^{\circ} C

2

1473^{\circ} C

3

927^{\circ} C

4

546^{\circ} C

Explanation:

C Given,
Initial temperature $= 27^{\circ} C = 300 K$
Initial velocity $= V$
Final velocity $= 2 V$
We know that,
$V_{rms} = \sqrt{\frac{3 R RT}{M}}$
$V \propto T^{1 / 2}$
$\frac{V_{1}}{T_{1}^{1 / 2}} = \frac{V_{2}}{T_{2}^{1 / 2}}$
$\frac{V}{(300)^{1 / 2}} = \frac{2 V}{{(T_{2})}^{1 / 2}}$
${(\frac{T_{2}}{300})}^{1 / 2} = 2$
$\frac{T_{2}}{300} = 4$
$T_{2} = 1200 K$
$T_{2} = 1200 - 273$
$T_{2} = 927^{\circ} C$
$∴ T_{2} = 1200 - 273$

Shift-II

Kinetic Theory of Gases

139121 The r.m.s. velocity of hydrogen molecules at temperature $T$ is seven times the r.m.s. velocity of nitrogen molecules at $300 K$ . This temperature $T$ is (Molecular weights of hydrogen and nitrogen are 2 and 28 respectively)

1

1350 K

2

1700 K

3

1050 K

4

2100 K

Explanation:

C We know that,
$V_{rms} = \sqrt{\frac{3 RT}{M}}$
${(V_{rms})}_{H_{2}} = 7 \times {(V_{rms})}_{N_{2}}$
$\sqrt{\frac{3 {RT}_{H_{2}}}{M_{H_{2}}}} = 7 \times \sqrt{\frac{3 {RT}_{N_{2}}}{M_{N_{2}}}}$
$∴ \frac{T_{H_{2}}}{2} = 49 \times \frac{T_{N_{2}}}{28}$
$T_{H_{2}} = 7 \times \frac{T_{N_{2}}}{2}$
$= 7 \times \frac{300}{2}$
$= 7 \times 150$
$= 1050 K$

MHT-CET 2020

Question Bank Series

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