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Let A and B the two gases and given : $ { T_A \over T_B } = 4 { M_A \over M_B } $ Where T is the temperature and M is molecular mass. If $ \nu_A $ and $ \nu_ B $ are the r.m.s speed, then the ratio $ { \nu_A \over \nu_B} $ will be equal to

A. 2 B. 4 C. 0.5 D. 1

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Explanation

${T_A \over M_A} = 4 { T_B \over M_B} \Rightarrow \sqrt { T_A \over M_A} = 2 \sqrt { T_B \over M_B}$ $ \Rightarrow \sqrt { 3RT_A \over M_A} = 2 \sqrt { 3RT_B \over M_B} \Rightarrow \nu_A = 2 \nu_B \Rightarrow { \nu_A \over \nu_B} = 2 $

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The rms. speed of the molecules of a gas in a vessel is $ 400 ms^{-1}$ . If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be

A. $ 800 ms^{-1} $ B. $ 200 ms^{-1} $ C. $ 400 \sqrt 2 ms^{-1} $ D. $ 400 ms^{-1} $

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Explanation

Since temperature is constant So $ \nu_{rms} $ remains same

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At which temperature the velocity of $O_2$ molecules will be equal to the rms velocity of $N_2$ molecules at $ 0 ^\circ C$

A. $ 93 ^\circ C$ B. $ 40 ^\circ C$ C. $ 39 ^\circ C$ D. Can't be determined

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Explanation

$ \nu_{rms} = \sqrt { 3RT \over M_o} \Rightarrow T \alpha M_o $ $( \nu_{rms} , R \rightarrow const )$ $ \Rightarrow { T_{O_2} \over T_{N_2} } = { (M_o)_{O_2} \over (M_o)_{N_2} }$

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The rms speed of the molecules of a gas at a pressure $10^5$ Pa and temperature $0 ^\circ C$ is 0.5 km/s. If the pressure is kept constant but temperature is raised to $ 819 ^\circ C$ , the rms speed becomes

A. $ 1.5 kms^{-1} $ B. $ 2 kms^{-1} $ C. $ 1 kms^{-1} $ D. $ 5 kms^{-1} $

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Explanation

$ \nu_{rms} \alpha \sqrt T $ $ \therefore { ( \nu_{rms} )_1 \over (\nu_{rms} )} = \sqrt { T_1 \over T_2} $

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The root mean square velocity of a gas molecule of mass m at a given temperature is proportional to

A. $ m ^\circ C$ B. $ m^{-1/2} $ C. $ m^{1/2} $ D. m

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Explanation

$ \nu_{rms} = \sqrt { 3 k_B T \over m } $ $ \therefore \nu_{rms} \alpha m^{-1/2 } $

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The ratio of the vapour densities of two gases at a given temperature is 9:8, The ratio of the rms velocities of their molecule is

A. $ 3 : 2 \sqrt 2 $ B. $ 2 \sqrt 2 : 3 $ C. 9:8 D. 8:9

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Explanation

At a given temperature $ \nu_{rms} \alpha { 1 \over \sqrt \rho } $

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At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value at $ O ^\circ C$ ?

A. 204.75 K B. $ 204 .75 ^\circ C$ C. -204.75 K D. $ '-204 .75 ^\circ C$

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Explanation

$ \nu_{rms } \alpha \sqrt T and ( \nu_{rms}) _2 = {1 \over 2 } ( \nu_{rms} )_1 $ $ \therefore { (\nu_{rms} )_2 \over (\nu_{rms})_1 } = \sqrt { T \over T_o} = {1 \over 2 } $ $ \therefore \sqrt { 273 + t \over 273 +0 } = {1 \over 2 } $ $ \therefore { 273 + t \over 273 } = {1 \over 4 } \Rightarrow t = { 273 \over 4 } -273 = 68.25 -273 = -204.75 ^\circ C $

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The rms velocity of gas molecules is $ 300 ms^{-1}$. The rms velocity of molecules of gas with twice the molecular weight and half the absolute temperature is

A. $ 300 ms^{-1} $ B. $ 150 ms^{-1} $ C. $ 600 ms^{-1} $ D. $ 75 ms^{-1} $

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Explanation

$ \nu_{rms} = \sqrt { 3RT \over Mo} $ $ (\nu_{rms} )_1= \sqrt { 3RT \over M_o} ; (\nu_{rms})_2 = \sqrt { 3R(T/2) \over 2M_o} = \sqrt { 3RT \over M_o } = { 1 \over 2 } \sqrt { 3RT \over 4M_o } = { (\nu_{rms} )_1 \over 2 } =150 ms^{-1} $

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Calculate the temperature at which rms velocity of $ S0_2$ molecules is the same as that of $O_2$ molecules at $ 27 ^\circ C$ . Molecular weights of Oxygen and $SO_2$ are 32 g and 64 g respectively

A. $ 327 ^\circ C$ B. 327 K C. $ 127 ^\circ C$ D. $ 227 ^\circ C$

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Explanation

$ \nu_{rms} = \sqrt { 3RT \over Mo} $ $ \left ( \nu_{rms} = \sqrt { 3RT \over Mo} = \sqrt { (3) (8.314) (290) \over 28 \times 10^{-3} } = 508.24 ms^{-1} \right) $ here $ (\nu_{rms} )_1 = (\nu_{rms})_2 $ $ \therefore { T_{O_2} \over (M_o)_{O_2 } } = { T_{SO_2} \over (M_o)_{SO_2 } } \Rightarrow { 300 \over 32 } = { T_{SO _2 } \over 64 } = 600 K = 327 ^\circ C $

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For a gas, the rms speed at 800 K is

A. Four times the value at 200 K B. Twice the value at 200 K C. Half the value at 200 K D. same as at 200 K

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Explanation

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

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