Practice free Kinetic Theory of Gases (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
For a gas, the rms speed at 800 K is
$ \nu_{rms} \alpha \sqrt T $ $ \therefore { \nu_1 \over \nu_2 } = \sqrt { T_1 \over T_2 } $
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
$ { (\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 $
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
$ \nu_{rms} \alpha \sqrt T \Rightarrow {\nu_2 \over \nu_1} = \sqrt { T_2 \over T_1} $
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 } $
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Kinetic Theory of Gases question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.