Internal energy and enthalpy of ideal gases depend only on temperature

Previously, we said that the enthalpy of an ideal gas is independent of pressure at constant temperature. And the internal energy of an ideal gas is independent of volume at constant temperature. The enthalpy and internal energy of an ideal gas were asserted to be functions of temperature only. Here, we can prove it using the property relations. Let’s begin with enthalpy. dH is TdS + VdP. Divide both side by dP at constant temperature. Then (dH over dP) is T times (dS over dP) + V. So, to get the pressure dependence of enthalpy at constant temperature, we need (dS over dP) at constant temperature. Among the property relations, it can be obtained from dG, since G has variables of P and T.

(dS over dP) at constant T is - (dV over dT) at constant P. So, (dH over dP) is now - T (dV over dT) + V. For ideal gas, V equals to RT over P from the equation of state. Then, (dH over dP) at constant T becomes zero. Thus, enthalpy does not depend on pressure at constant T and it is a function of temperature only. Similarly, let’s prove that the internal energy of an ideal gas is a function of temperature only and independent of volume. dU is TdS - PdV. Divide both sides with dV at constant T. Then the volume dependence of the internal energy can be calculated from (dS over dV) at constant T.

This can be obtained from dF since F has T, V as variables. dF is - SdT - PdV. The property relation from here is (dS over dV) at constant T equals (dP over dT) at constant V. So, (dU over dV) is T times (dP over dV) - P. Again P is RT over V for the ideal gas. Thus (dU over dV) at constant T is zero. So the internal energy is a function of temperature only.