Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
At which temperature the velocity of $O_2$ molecules will be equal to the rms velocity of $N_2$ molecules at $ 0 ^\circ C$
$ \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} }$
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
$ \nu_{rms} \alpha \sqrt T $ $ \therefore { ( \nu_{rms} )_1 \over (\nu_{rms} )} = \sqrt { T_1 \over T_2} $
The root mean square velocity of a gas molecule of mass m at a given temperature is proportional to
$ \nu_{rms} = \sqrt { 3 k_B T \over m } $ $ \therefore \nu_{rms} \alpha m^{-1/2 } $
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
At a given temperature $ \nu_{rms} \alpha { 1 \over \sqrt \rho } $
At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value at $ O ^\circ C$ ?
$ \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 $
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
$ \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} $
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
$ \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 $
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} $
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every NEET 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.