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Which of the following statement is true?

Absolute zero degree temperature is not zero energy temperature.

Two different gases at the same temperature pressure have equal root mean square velocities.

The root mean square speed of the molecules of different ideal gases, maintained at the same

temperature are the same.

Given sample of 1 cc of hydrogen and 1 cc of oxygen both at NTP; oxygen sample has a large

the number of molecules.

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Explanation

1. Absolute zero degree temperature is not zero energy temperature.

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At which temperature the velocity of $O_{2}$ molecules will be equal to the velocity of $N_{2}$ molecules at 0°C

40°C

93°C

39°C

Cannot be calculated

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Explanation

$3 . v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow T \propto M R \to constant \Rightarrow \frac{T_{O_{2}}}{T_{H_{2}}} = \frac{M_{O_{2}}}{M_{H_{2}}} \Rightarrow \frac{T_{O_{2}}}{(273 + 0)} = \frac{32}{28} \Rightarrow T_{O_{2}} = 312 K = 39 ° C$

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The respective speeds of the molecules are 1, 2, 3, 4 and 5 km/sec. The ratio of their r.m.s. velocity and the average velocity will be

$\sqrt{11} : 3$

$3 : \sqrt{11}$

$1 : 2$

$3 : 4$

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Explanation

$1 . v_{rms} = \sqrt{\frac{1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2}}{5}} = \sqrt{11} km / s v_{av} = \frac{1 + 2 + 3 + 4 + 5}{5} = 3 km / s \Rightarrow \frac{v_{rms}}{v_{av}} = \frac{\sqrt{11}}{3}$

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According to the kinetic theory of gases the r.m.s. velocity of gas molecules is directly proportional to

$T$

$\sqrt{T}$

$T^{2}$

$1 / \sqrt{T}$

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Explanation

2. $v_{rms} \propto \sqrt{T}$

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At a given temperature the root mean square velocities of oxygen and hydrogen molecules are in the ratio

16 : 1

1 : 16

4 : 1

1 : 4

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Explanation

$4 . v_{rms} \propto \frac{1}{\sqrt{M}}; so \frac{(v_{rms}) o_{2}}{{(v_{rms})}_{H}_{2}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{2}{32}} = 1 : 4$

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The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square velocity of the gas molecules is v, at 480 K it becomes

v/2

v/4

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Explanation

$2 . v_{rms} \propto \sqrt{T} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{120}{480}} = \frac{1}{2} \Rightarrow v_{2} = 2 v$

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By what factor the r.m.s. velocity will change, if the temperature is raised from 27°C to 327°C

$\sqrt{2}$

$2$

$2 \sqrt{2}$

$1$

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Explanation

1. $v_{rms} \propto \sqrt{T} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{(273 + 27)}{(273 + 327)}} = \sqrt{\frac{300}{600}} = \frac{1}{\sqrt{2}} \Rightarrow v_{2} = \sqrt{2} v_{1}$

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At a given temperature if $V_{rms}$ is the root mean square velocity of the molecules of a gas and $V_{s}$ the velocity of sound in it, then these are related as $(γ = \frac{C_{P}}{C_{v}})$

$V_{rms} = V_{s}$

$V_{rms} = \sqrt{\frac{3}{γ}} \times V_{s}$

$V_{rms} = \sqrt{\frac{γ}{3}} \times V_{s}$

$V_{rms} = (\frac{3}{γ}) \times V_{s}$

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Explanation

$(2) V_{rms} = \sqrt{\frac{3 P}{ρ}}, and V_{sound} = \sqrt{\frac{γP}{ρ}} So \frac{V_{rms}}{V_{sound}} = \sqrt{\frac{3}{γ}} \Rightarrow V_{rms} = \sqrt{\frac{3}{γ}} \times V_{sound}$

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Let A and B the two gases and given : $\frac{T_{A}}{M_{A}} = 4 . \frac{T_{B}}{M_{B}};$ where T is the temperature and M is molecular mass. If $C_{A}$ and $C_{B}$ are the r.m.s. speed, then the ratio $\frac{C_{A}}{C_{B}}$ will be equal to

0.5

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Explanation

$1 . \frac{T_{A}}{M_{A}} = 4 . \frac{T_{B}}{M_{B}} \Rightarrow \sqrt{\frac{T_{A}}{M_{A}}} = 2 \sqrt{\frac{T_{B}}{M_{B}}} \Rightarrow \sqrt{\frac{3 {RT}_{A}}{M_{A}}} = 2 \sqrt{\frac{3 RT}{M_{B}}} \Rightarrow C_{A} = 2 C_{B} \Rightarrow \frac{C_{A}}{C_{B}} = 2$

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For a gas at a temperature the root-mean-square velocity $v_{rms}$ , the most probable speed $v_{mp}$ , and the average speed $v_{av}$ obey the relationship

$v_{av} > v_{rms} > v_{mp}$

$v_{rms} > v_{av} > v_{mp}$

$v_{mp} > v_{av} > v_{rms}$

$v_{mp} > v_{rms} > v_{av}$

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Explanation

2. $v_{rms} > v_{av} > v_{mp}$

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