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At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are in the ratio
$ \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}} $
If mass of He atom is 4 times that of hydrogen atom, then rms speed of the is
$ \nu_{rms} \alpha \sqrt { 1 \over m } $ $ \therefore { \nu_{He} \over \nu_{H} } = \sqrt { m_H \over m_{He} } $
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
$ \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} }$
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
$ \nu_{rms} = \sqrt { 3P \over \rho} or P = { \rho \nu^2_{rms} \over 3 } $
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
$ \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 \% $
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
$ \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 $CH_4$
The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earth surface will be
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 } $
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} $
$ \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) $
The radius of a molecule of Argon gas is $ 1.78 A ^\circ C$ . Find the mean free path of molecules of Argon at 0° C temperature and 1 atm pressure. $ k_B = 1.38 \times 10^{-23} JK^{-1} $
$ \bar l = { 1 \over \sqrt 2 \pi n d^2 } = {k_B T \over \sqrt 2 \pi Pd^2 } = 6.65 \times 10^{-8} m $
A monoatomic gas molecule has
A monoatomic gas moecule has only 3 translational degrees of freedom
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