Physics · Kinetic Theory · NEET
No. For an ideal gas, U depends only on the absolute temperature T. The formula U = (f/2) nRT has no P or V in it. If you compress a gas but keep T the same (isothermal), U does not change. This is why in an isothermal process the change in internal energy is zero. Pressure and volume can change, but as long as T is fixed, U is fixed.
An ideal gas is assumed to have no forces between molecules except during collisions. Internal energy is the sum of molecular kinetic energy plus molecular potential energy. Since there are no intermolecular forces, the potential energy part is zero. So only kinetic energy is left, and kinetic energy depends only on temperature. That is why U is a function of T alone for an ideal gas.
A monatomic gas (like He, Ar) has f = 3 (three translational degrees of freedom), so U = (3/2) nRT. A diatomic gas (like O2, N2, H2) treated as a rigid rotator has f = 5 (3 translational + 2 rotational), so U = (5/2) nRT. A polyatomic gas has f = 6 (3 translational + 3 rotational) for the rigid case, giving U = 3 nRT. More degrees of freedom means more internal energy at the same temperature.
Equipartition says each degree of freedom carries an average energy of (1/2) kB T per molecule. A molecule with f degrees of freedom has average energy (f/2) kB T. One mole has NA molecules, so energy per mole is (f/2) NA kB T = (f/2) RT, because NA kB = R. For n moles, U = (f/2) nRT.
Since U = (f/2) nRT, the change is ΔU = (f/2) nR ΔT. This depends only on the temperature change, not on the path taken. Whether the process is isobaric, isochoric or any other, if the temperature rises by the same ΔT, ΔU is the same. For example, for a monatomic gas ΔU = (3/2) nR ΔT.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
U = (f/2) nRT, where f is degrees of freedom, n is moles, R is the gas constant and T is absolute temperature in kelvin.
Yes. It depends only on the current state (temperature), not on how the gas reached that state. That is why ΔU over a cyclic process is zero.
For one mole of a monatomic gas, U = (3/2) RT, since it has 3 translational degrees of freedom.
No. An ideal gas has no intermolecular forces, so molecular potential energy is zero. The internal energy is entirely molecular kinetic energy.
Since ΔU = (f/2) nR ΔT and ΔU = n Cv ΔT at constant volume, we get Cv = (f/2) R. This links internal energy directly to the molar specific heat Cv.
