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PHXI13:KINETIC THEORY

360225 If the $r m s$ speed, of molecules of an ideal gas is ' $v$ ', what is the average speed of the molecules at the same temperature

1

\sqrt{\frac{8}{3 π}} v

2

\sqrt{\frac{2}{3}} v

3

v

4

\frac{3 v}{2}

Explanation:

$v_{r m s} = \sqrt{3 R T / M}, V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

PHXI13:KINETIC THEORY

360226 The temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$ , then percentage increase in $v_{r m s}$ is

1

11 %

2

33 %

3

37 %

4

15.5 %

Explanation:

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$% increases in v_{r m s} = \frac{\sqrt{\frac{3 R T_{2}}{M}} - \sqrt{\frac{3 R T_{1}}{M}}}{\sqrt{\frac{3 R T_{1}}{M}}} \times 100$
$= \frac{20 - 17.32}{17.32} \times 100 = 15.5$

PHXI13:KINETIC THEORY

360227 A mixture of 2 moles of helium gas $(a t o m i c m a s s = 4 u)$ , and 1 mole of argon gas $(a t o m i c m a s s = 40 u)$ is kept at $300 K$ in a container. The ratio of their rms speeds $[\frac{v_{mm} (helium)}{v_{rms} (argon)}]$ is close to

1 2.24

2 0.32

3 0.45

4 3.16

Explanation:

$v_{m s} = \sqrt{\frac{3 R T}{m}}$
$\frac{{(v_{rms})}_{He}}{{(v_{rms})}_{Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$

JEE - 2019

PHXI13:KINETIC THEORY

360228 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule $(m) = 2.76 \times 10^{- 26} k g$ , Boltzmann's constant $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$

1

5.016 \times 10^{4} K

2

1.254 \times 10^{4} K

3

83.6 \times 10^{3} K

4

2.508 \times 10^{4} K

Explanation:

$v_{r m s} = v_{e s c a p e}$
$\sqrt{\frac{3 K T}{m}} = 11.2 km / s$
$\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}} = {(11.2 \times 10^{3})}^{2};$
$T = \frac{2.76 \times 125.44 \times 10^{6}}{3 \times 1.38 \times 10^{3}} = 83.6 \times 10^{3} K$

NEET - 2018

PHXI13:KINETIC THEORY

360229 At room temperature a diatomic gas is found to have an r.m.s speed of $1930 m s^{- 1}$ . The gas is

1

H_{2}

2

C l_{2}

3

O_{2}

4

F_{2}

Explanation:

$∵ c = \sqrt{\frac{3 R T}{M}}$
$(1930)^{2} = \frac{3 \times 8.314 \times 300}{M}$
$M = \frac{3 \times 8.314 \times 300}{1930 \times 1930} \approx 2 g$
The gas is $H_{2}$

JEE - 2014

PHXI13:KINETIC THEORY

360225 If the $r m s$ speed, of molecules of an ideal gas is ' $v$ ', what is the average speed of the molecules at the same temperature

1

\sqrt{\frac{8}{3 π}} v

2

\sqrt{\frac{2}{3}} v

3

v

4

\frac{3 v}{2}

Explanation:

$v_{r m s} = \sqrt{3 R T / M}, V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

PHXI13:KINETIC THEORY

360226 The temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$ , then percentage increase in $v_{r m s}$ is

1

11 %

2

33 %

3

37 %

4

15.5 %

Explanation:

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$% increases in v_{r m s} = \frac{\sqrt{\frac{3 R T_{2}}{M}} - \sqrt{\frac{3 R T_{1}}{M}}}{\sqrt{\frac{3 R T_{1}}{M}}} \times 100$
$= \frac{20 - 17.32}{17.32} \times 100 = 15.5$

PHXI13:KINETIC THEORY

360227 A mixture of 2 moles of helium gas $(a t o m i c m a s s = 4 u)$ , and 1 mole of argon gas $(a t o m i c m a s s = 40 u)$ is kept at $300 K$ in a container. The ratio of their rms speeds $[\frac{v_{mm} (helium)}{v_{rms} (argon)}]$ is close to

1 2.24

2 0.32

3 0.45

4 3.16

Explanation:

$v_{m s} = \sqrt{\frac{3 R T}{m}}$
$\frac{{(v_{rms})}_{He}}{{(v_{rms})}_{Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$

JEE - 2019

PHXI13:KINETIC THEORY

360228 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule $(m) = 2.76 \times 10^{- 26} k g$ , Boltzmann's constant $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$

1

5.016 \times 10^{4} K

2

1.254 \times 10^{4} K

3

83.6 \times 10^{3} K

4

2.508 \times 10^{4} K

Explanation:

$v_{r m s} = v_{e s c a p e}$
$\sqrt{\frac{3 K T}{m}} = 11.2 km / s$
$\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}} = {(11.2 \times 10^{3})}^{2};$
$T = \frac{2.76 \times 125.44 \times 10^{6}}{3 \times 1.38 \times 10^{3}} = 83.6 \times 10^{3} K$

NEET - 2018

PHXI13:KINETIC THEORY

360229 At room temperature a diatomic gas is found to have an r.m.s speed of $1930 m s^{- 1}$ . The gas is

1

H_{2}

2

C l_{2}

3

O_{2}

4

F_{2}

Explanation:

$∵ c = \sqrt{\frac{3 R T}{M}}$
$(1930)^{2} = \frac{3 \times 8.314 \times 300}{M}$
$M = \frac{3 \times 8.314 \times 300}{1930 \times 1930} \approx 2 g$
The gas is $H_{2}$

JEE - 2014

PHXI13:KINETIC THEORY

360225 If the $r m s$ speed, of molecules of an ideal gas is ' $v$ ', what is the average speed of the molecules at the same temperature

1

\sqrt{\frac{8}{3 π}} v

2

\sqrt{\frac{2}{3}} v

3

v

4

\frac{3 v}{2}

Explanation:

$v_{r m s} = \sqrt{3 R T / M}, V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

PHXI13:KINETIC THEORY

360226 The temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$ , then percentage increase in $v_{r m s}$ is

1

11 %

2

33 %

3

37 %

4

15.5 %

Explanation:

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$% increases in v_{r m s} = \frac{\sqrt{\frac{3 R T_{2}}{M}} - \sqrt{\frac{3 R T_{1}}{M}}}{\sqrt{\frac{3 R T_{1}}{M}}} \times 100$
$= \frac{20 - 17.32}{17.32} \times 100 = 15.5$

PHXI13:KINETIC THEORY

360227 A mixture of 2 moles of helium gas $(a t o m i c m a s s = 4 u)$ , and 1 mole of argon gas $(a t o m i c m a s s = 40 u)$ is kept at $300 K$ in a container. The ratio of their rms speeds $[\frac{v_{mm} (helium)}{v_{rms} (argon)}]$ is close to

1 2.24

2 0.32

3 0.45

4 3.16

Explanation:

$v_{m s} = \sqrt{\frac{3 R T}{m}}$
$\frac{{(v_{rms})}_{He}}{{(v_{rms})}_{Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$

JEE - 2019

PHXI13:KINETIC THEORY

360228 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule $(m) = 2.76 \times 10^{- 26} k g$ , Boltzmann's constant $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$

1

5.016 \times 10^{4} K

2

1.254 \times 10^{4} K

3

83.6 \times 10^{3} K

4

2.508 \times 10^{4} K

Explanation:

$v_{r m s} = v_{e s c a p e}$
$\sqrt{\frac{3 K T}{m}} = 11.2 km / s$
$\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}} = {(11.2 \times 10^{3})}^{2};$
$T = \frac{2.76 \times 125.44 \times 10^{6}}{3 \times 1.38 \times 10^{3}} = 83.6 \times 10^{3} K$

NEET - 2018

PHXI13:KINETIC THEORY

360229 At room temperature a diatomic gas is found to have an r.m.s speed of $1930 m s^{- 1}$ . The gas is

1

H_{2}

2

C l_{2}

3

O_{2}

4

F_{2}

Explanation:

$∵ c = \sqrt{\frac{3 R T}{M}}$
$(1930)^{2} = \frac{3 \times 8.314 \times 300}{M}$
$M = \frac{3 \times 8.314 \times 300}{1930 \times 1930} \approx 2 g$
The gas is $H_{2}$

JEE - 2014

PHXI13:KINETIC THEORY

360225 If the $r m s$ speed, of molecules of an ideal gas is ' $v$ ', what is the average speed of the molecules at the same temperature

1

\sqrt{\frac{8}{3 π}} v

2

\sqrt{\frac{2}{3}} v

3

v

4

\frac{3 v}{2}

Explanation:

$v_{r m s} = \sqrt{3 R T / M}, V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

PHXI13:KINETIC THEORY

360226 The temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$ , then percentage increase in $v_{r m s}$ is

1

11 %

2

33 %

3

37 %

4

15.5 %

Explanation:

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$% increases in v_{r m s} = \frac{\sqrt{\frac{3 R T_{2}}{M}} - \sqrt{\frac{3 R T_{1}}{M}}}{\sqrt{\frac{3 R T_{1}}{M}}} \times 100$
$= \frac{20 - 17.32}{17.32} \times 100 = 15.5$

PHXI13:KINETIC THEORY

360227 A mixture of 2 moles of helium gas $(a t o m i c m a s s = 4 u)$ , and 1 mole of argon gas $(a t o m i c m a s s = 40 u)$ is kept at $300 K$ in a container. The ratio of their rms speeds $[\frac{v_{mm} (helium)}{v_{rms} (argon)}]$ is close to

1 2.24

2 0.32

3 0.45

4 3.16

Explanation:

$v_{m s} = \sqrt{\frac{3 R T}{m}}$
$\frac{{(v_{rms})}_{He}}{{(v_{rms})}_{Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$

JEE - 2019

PHXI13:KINETIC THEORY

360228 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule $(m) = 2.76 \times 10^{- 26} k g$ , Boltzmann's constant $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$

1

5.016 \times 10^{4} K

2

1.254 \times 10^{4} K

3

83.6 \times 10^{3} K

4

2.508 \times 10^{4} K

Explanation:

$v_{r m s} = v_{e s c a p e}$
$\sqrt{\frac{3 K T}{m}} = 11.2 km / s$
$\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}} = {(11.2 \times 10^{3})}^{2};$
$T = \frac{2.76 \times 125.44 \times 10^{6}}{3 \times 1.38 \times 10^{3}} = 83.6 \times 10^{3} K$

NEET - 2018

PHXI13:KINETIC THEORY

360229 At room temperature a diatomic gas is found to have an r.m.s speed of $1930 m s^{- 1}$ . The gas is

1

H_{2}

2

C l_{2}

3

O_{2}

4

F_{2}

Explanation:

$∵ c = \sqrt{\frac{3 R T}{M}}$
$(1930)^{2} = \frac{3 \times 8.314 \times 300}{M}$
$M = \frac{3 \times 8.314 \times 300}{1930 \times 1930} \approx 2 g$
The gas is $H_{2}$

JEE - 2014

PHXI13:KINETIC THEORY

360225 If the $r m s$ speed, of molecules of an ideal gas is ' $v$ ', what is the average speed of the molecules at the same temperature

1

\sqrt{\frac{8}{3 π}} v

2

\sqrt{\frac{2}{3}} v

3

v

4

\frac{3 v}{2}

Explanation:

$v_{r m s} = \sqrt{3 R T / M}, V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

PHXI13:KINETIC THEORY

360226 The temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$ , then percentage increase in $v_{r m s}$ is

1

11 %

2

33 %

3

37 %

4

15.5 %

Explanation:

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$% increases in v_{r m s} = \frac{\sqrt{\frac{3 R T_{2}}{M}} - \sqrt{\frac{3 R T_{1}}{M}}}{\sqrt{\frac{3 R T_{1}}{M}}} \times 100$
$= \frac{20 - 17.32}{17.32} \times 100 = 15.5$

PHXI13:KINETIC THEORY

360227 A mixture of 2 moles of helium gas $(a t o m i c m a s s = 4 u)$ , and 1 mole of argon gas $(a t o m i c m a s s = 40 u)$ is kept at $300 K$ in a container. The ratio of their rms speeds $[\frac{v_{mm} (helium)}{v_{rms} (argon)}]$ is close to

1 2.24

2 0.32

3 0.45

4 3.16

Explanation:

$v_{m s} = \sqrt{\frac{3 R T}{m}}$
$\frac{{(v_{rms})}_{He}}{{(v_{rms})}_{Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$

JEE - 2019

PHXI13:KINETIC THEORY

360228 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule $(m) = 2.76 \times 10^{- 26} k g$ , Boltzmann's constant $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$

1

5.016 \times 10^{4} K

2

1.254 \times 10^{4} K

3

83.6 \times 10^{3} K

4

2.508 \times 10^{4} K

Explanation:

$v_{r m s} = v_{e s c a p e}$
$\sqrt{\frac{3 K T}{m}} = 11.2 km / s$
$\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}} = {(11.2 \times 10^{3})}^{2};$
$T = \frac{2.76 \times 125.44 \times 10^{6}}{3 \times 1.38 \times 10^{3}} = 83.6 \times 10^{3} K$

NEET - 2018

PHXI13:KINETIC THEORY

360229 At room temperature a diatomic gas is found to have an r.m.s speed of $1930 m s^{- 1}$ . The gas is

1

H_{2}

2

C l_{2}

3

O_{2}

4

F_{2}

Explanation:

$∵ c = \sqrt{\frac{3 R T}{M}}$
$(1930)^{2} = \frac{3 \times 8.314 \times 300}{M}$
$M = \frac{3 \times 8.314 \times 300}{1930 \times 1930} \approx 2 g$
The gas is $H_{2}$

JEE - 2014