MHT CET202619 April 2026Evening ShiftPhysicsKinetic Theory of GasesActual
Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is
Options
- A( 2 3 )^ rd of kinetic energy per unit volume of a gas.
- B( 3 4 )^ th of kinetic energy per unit volume of a gas.
- C( 1 3 )^ rd of kinetic energy per unit volume of a gas.
- D( 3 2 )^ nd times kinetic energy per unit volume of a gas.
Correct answer
A. ( 2 3 )^ rd of kinetic energy per unit volume of a gas.
Step-by-step solution
The pressure exerted by an ideal gas is given by P = 1 3 v_ rms ^2 where is the density of the gas and v_ rms is the root mean square velocity. The translational kinetic energy per unit volume of the gas is E = 1 2 v_ rms ^2 Multiplying and dividing the pressure equation by 2 , we get P = 2 3 ( 1 2 v_ rms ^2 ) P = 2 3 E Thus, the pressure is ( 2 3 )^ rd of the kinetic energy per unit volume of the gas. Answer: ( 2 3 )^ rd of kinetic energy per unit volume of a gas.