Practice free Kinetic Theory of Gases (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
If the ratio of vapour density for hydrogen and oxygen is 1/ 16 , then under constant pressure the ratio of their rms velocities will be
$ \nu _ {rms} =\sqrt { 3P \over \rho } \Rightarrow { \nu_1 \over \nu_2 } = \sqrt { \rho_2 \over \rho_1} = \sqrt { 16 \over 1} = 4:1 $
The molecules of a given mass of a gas have a rms velocity of 200 m /s at $ 27 ^\circ C$ and $ 1.0 \times 10^5 Nm^{-2} $ pressure when the temperature is $ 127 ^\circ C$ and pressure is $ 0.5 \times 10^5 Nm^{-2} $ , the rms velocity in m s will be
rms velocity doesn't depend on pressure, it depends upon temperature only $ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu_{rms} \alpha \sqrt T $ $ \therefore T \alpha \nu_{rms}^2 \Rightarrow { \nu_1 \over \nu_2 } = \sqrt { T_1 \over T_2 } $
If the molecular weight of two gases are $ M_1$ and $M_2$ , then at a given temperature the ratio of root mean square velocity $ \nu _1 $ and $ \nu _2 $ will be
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu_1 \alpha \sqrt { 1 \over M_1 } and \nu_2 \alpha \sqrt { 1 \over M_2} \Rightarrow { \nu_1 \over \nu_2 } = \sqrt { M_2 \over M_1 } $
To what temperature should the hydrogen at $ 327 ^\circ C$ cooled at constant pressure, so that the root mean square velocity of its molcules become half of its previous value
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu_{rms} \alpha \sqrt T $ $ \therefore T \alpha \nu_{rms} ^2 $
At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at $ 47 ^\circ C$ ?
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow T \alpha M_o $ (\nu_{rms} . R \rightarrow constant ) $
The root mean square velocity of the molecules in a sample of helium is 5/7 th that of the molecules in a sample of hydrogen. If the temperature of hydorgen sample is $0 ^\circ C$ , then the temperature of the helium sample is about
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu _{rms} \alpha \sqrt { T \over Mo} $ $ \therefore { \nu_{He} \over \nu_{H_2}} ={ 5 \over 7} = \sqrt { T_{He} \times M_{H_2} \over T_{H2} M_{He} } \Rightarrow = { 25 \over 49 } \times {4 \over 2 } \times 273 $ $ \approx 273 K \approx 0 ^ \circ C$
At room temperature ( $ 27 ^\circ C$ ), the rms speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow M_o = { 3RT \over \nu_{rms} ^2 } = { 3 \times 8.3 \times 300 \over (1920)^2} = 2 \times 10^{-3} kg = 2g $ Gas is hydrogen
If three molecules have velocities 0.5, 1 and 2 the ratio of rms speed and average speed is (The velocities are in km/s)
rms speed, $ \nu_{rms} = \sqrt { \nu_1^2 +\nu_2^2 + \nu_3^2 \over 3 }$ average speed .$ (\nu) = { \nu_1 +\nu_2 + \nu_3 \over 3 }$
At what temperature pressure remaining constant will the rms speed of a gas molecules increase by 10% of the rms speed at NTP?
$ { ( \nu_{rms} )_t \over ( \nu_{rms})_o} = \sqrt { T \over T_o} and rms speed increases by 10 \% $
$ \therefore 1.1 = \sqrt { 273 + t \over 273} or {273 + t \over 273 } = (1.1)^2 = 1.21 $
$ \therefore t = 273 ( 1.21 - 1 ) = 57.3 C$
When temperature of an ideal gas is increased from $ 27 ^\circ C$ to $ 227 ^\circ C$ , its rms speed changed from $400 ms^{–1}$ to V . The V is
$ \nu_{rms} \alpha \sqrt T $ $ \therefore { \nu_2 \over \nu_1} = \sqrt { T_2 \over T_1} $
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