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A mixture of 2 moles of helium gas (atomic mass = 4 amu), and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds
$ { \nu_{rms} (helium) \over \nu_{rms} (argon) $ is

A. 0.45 B. 2.24 C. 3.16 D. 0.32

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Explanation

$ { (\nu _{rms} )_{He} \over (\nu_{rms} )_{Ar} } = { \sqrt { 3RT \over (M_o)_{He}} \over \sqrt { 3RT \over (M_o)_{Ar}}}= \sqrt { 40 \over 4 } = \sqrt 10 \approx 3.16 $

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The temperature of an ideal gas is increased from $ 27 ^\circ C$ to $ 927 ^\circ C$ . The root mean square speed of its molecules becomes

A. Four times B. One-fourth C. Half D. Twice

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Explanation

$ \nu_{rms} \alpha \sqrt T \Rightarrow {\nu_2 \over \nu_1} = \sqrt { T_2 \over T_1} $

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

A. 1:4 B. 1:16 C. 16:1 D. 4:1

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Explanation

$ \nu_{rms} \sqrt { 1 \over M_o} $ So $ { (\nu_{rms} )_{O_2} \over (\nu_{rms} )_{H_2}} = \sqrt { (M_o)_{H_2} \over (M_o)_{O_2}} $

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If mass of He atom is 4 times that of hydrogen atom, then rms speed of the is

A. Two times of $ H - \nu_{rms} $ speed. B. Four times of $ H - \nu_{rms} $ speed. C. Same as of $ H - \nu_{rms} $ speed. D. Half of $ H - \nu_{rms} $ speed.

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Explanation

$ \nu_{rms} \alpha \sqrt { 1 \over m } $ $ \therefore { \nu_{He} \over \nu_{H} } = \sqrt { m_H \over m_{He} } $

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At temperature T, the rms speed of helium molecules is the same as rms speed of hydrogen mdecules at normal temperature and pressure. The value of T is

A. $ 546 ^\circ C$ B. $ 0 ^\circ C$ C. $ 273 ^\circ C$ D. $ 136.5 ^\circ C$

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Explanation

$ \nu_{rms} = \sqrt { 3RT \over M_o } \Rightarrow T \alpha M_o \Rightarrow {T_{He} \over T_H} = { (M_o)_{He} \over (M_o)_{H} }$

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If rms speed of a gas is $ \nu_ {rms} = 1840 m/s $ and its density $ \rho = 8.99 \times 10^{-2} kg/m3 $ , the pressure of the gas will be

A. $ 1.01 \times 10^3 Nm^{-2} $ B. $ 1.01 \times 10^5 Nm^{-2} $ C. $ 1.01 \times 10^7 Nm^{-2} $ D. $ 1.01 Nm^{-2} $

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Explanation

$ \nu_{rms} = \sqrt { 3P \over \rho} or P = { \rho \nu^2_{rms} \over 3 } $

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When the temperature of a gas is raised from $ 27 ^\circ C$ to $ 90 ^\circ C$ , the percentage increase in the rms velocity of the molecules will be

A. 15% B. 17.50% C. 10% D. 20%

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Explanation

$ \nu_{rms} = \sqrt { 3RT \over M_o} \Rightarrow {\nu_2 \over \nu_1} = \sqrt {T-2 \over T_1 } = \sqrt { 273+90 \over 273+27 } =1.1 $ $ \therefore \% increase = \left( {\nu_2 \over \nu_1} -1 \right) \times 100 \% = 0.1 \times 100 \% = 10 \% $

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The rms speed of a gas at a certain temperature is $ \sqrt 2 $ times than that of the Oxygen molecule at that temperature, the gas is

A. $ SO _2$ B. $ CH_4 $ C. $ H_2 $ D. He

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Explanation

$ \nu_{rms} \alpha \sqrt { 1 \over M_o} \Rightarrow {\nu_1 \over \nu_2} = \sqrt { (M_o)_2 \over (M_o)_1} $ $ \therefore {1 \over \sqrt 2} = \sqrt { (M-o)_2 \over 32} \Rightarrow (M_o)_2 = 16 $ Therefore the gas is $CH_4$

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The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earth surface will be

A. 5030 K B. 10063K C. 1060 K D. 8270 K

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Explanation

Escape velocity from the Earth's surface is $ 11.2 kms^{-1} $ So, $ \nu_{rms} = V _ {escape } = \sqrt { 3RT \over M_o } \Rightarrow T = { (\nu_{escape}) ^2 \times M_o \over 3R } $

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What is the meanfree path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm and temperature $ 17 ^\circ C $ ? Take the radius of nitrogen molecule to be 1A . Molecular mass of nitrogen = 28 , $ k_B = 1.38 \times 10^{-23} JK^{-1} , 1 atm = 1.013 \times 10^5 Nm^{-2} $

A. $ 2.2 \times 10^{-7} m, 2.58 \times 10^9 $ B. $ 1.1 \times 10^{-7} m, 4.58 \times 10^8 $ C. $ 1.1 \times 10^{-7} m, 4.58 \times 10^9 $ D. $ 2.2 \times 10^{-7} m, 3.58 \times 10^9 $

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Explanation

$ \bar l = { k_B T \over \sqrt 2 \pi P d^2 } = { (1.38 \times 10^{-23}) (290) \over (\sqrt 2 )(3.14) ( 2.026 \times 10^{-5} ) (2 \times 10^{-10} )^2 }= 1.11 \times 10^{-7} m $ Collision Frequency = no. of collision per second = $ { \nu_{rms} \over l} = { 508.24 \over 1.11 \times 10^{-7} } = 4.58 \times 10^9 $ $ \left( \nu_{rms} = \sqrt { 3RT \over M_o } = \sqrt { (3) (8.314) (290 ) \over 28 \times 10^{-3}} = 508.24 ms^{-1} \right) $

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