Practice free Kinetic Theory of Gases (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A vessel contains 1 mole of $O_2$ gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has a pressure of
$ PV = RT \Rightarrow P = T $ ( V and R = constant ) $ \Rightarrow { P_2 \over P_1 } = { T_2 \over T_1 } $
The pressure and temperature of two different gases P and T having the volumes V for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be,
When two gases with the same volume and temperature are mixed, their pressures add up. Since both gases have the same pressure P, the total pressure of the mixture will be 2P.
Air is filled at $ 60 ^\circ C$ in a vessel of open mouth. The vessel is heated to a temperature T so that 1/4 th part of air escapes. Assuming the volume of the vessel remaining constant the value of T is.
For open mouth vessel, pressure is constant. volume is also given constant. Hence from $ \therefore PV = RT $ $ PV = { M \over M_o } RT \Rightarrow T \alpha { 1 \over M } $ $ \therefore { T_1 \over T_2 } = {M_2 \over M_1 } $ 1/ 4 th part escapes, so remaining mass in the vessel is $ M_2 = {3 \over 4 } M_1 \Rightarrow { 273 + 60 \over T } = { { 3 \over 4 } M_1 \over M_1 } \Rightarrow T = 444 K = 171 C $
A gas is filled in a cylinder, its temperature is incresecd by 20% on kelvin scale and volume is reduced by 10%. How much percentage of the gas will leak out
Let initial conditions = V, T final conditions = V', T' By Charle's law, $ V \alpha T $ ( P remains constant ) But as per question, volume is reduced by 10% means Therefore V' = 0.9 V so percentage of volume leaked out $= { (1.2 - 0.9 ) V \over 1.2 V } \times 100 \% = 25 \% $
The pressure is exerted by the gas on the walls of the container because
$ Pressure P = { F \over A } = { 1 \over A } { \triangle P \over \triangle t } $ $ ( \triangle P = change in momentum) $
The relation between the gas pressure P and average kinetic energy per unit volume E is
$$ P = {2 \over 3 } ( Energy per unit volume ) $$ $$ = { 2 \over 3 } { E \over V } = { 2 \over 3 } E $$
The root mean square speed of hydrogen molecules of an ideal hydrogen kept in a gas chamber at $ 0 ^\circ C$ is $ 3180 ms^{–1} $. The pressure on the hydrogen gas is $ (Density of hydrogen gas is 8.99 \times 10^{-2} kg/ m3 , 1 atm = 1.01 \times 10^5 Nm^{-2 } ) $
$ V_{rms} = \sqrt { 3P \over \rho } \Rightarrow P = { V^2 _ {rms } \rho \over 3 } $
Gas at a pressure $P_o$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure will be equal to
$ V _{rms } = \sqrt { 3P \over \rho} = \sqrt { 3PV \over M } \Rightarrow V_{rms } \alpha \sqrt { P \over M } $ $ \therefore{ V_1 \over V_2 }= \sqrt { {P_1 \over P_2 } \times { M_2 \over M_1 }} $
A cylinder of capacity 20 litres is filled with $H_2$ gas. The total average kinetic energy of translatory motion of its molecules is $1.5 \times 10^5 J $. The pressure of hydrogen in the cylinder is
$ E = { 3 \over 2 } PV \Rightarrow P = { 2 E \over 3V } $
The average kinetic energy per molecule of a gas at $ -23 ^\circ C$ and 75 cm pressure is $ 5 \times 10^{-14} $ erg for $H_2$ . The mean kinetic energy per molecule of the $O_2$ at $ 227 ^\circ C $ and 150 cm pressure will be
The average kinetic energy $ E = { 3 \over 2 } k_B T $ $ \therefore { E_1 \over E_2 } = { T_1 \over T_2 } $
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.