Question

For a van der Waals gas, the internal energy U

• A. Depends only on Temperature
• B. Depends on both Temperature and Volume
• C. Is zero
• D. Depends only on Volume

Hint:

Keep these distinctions in mind for exams:
• Ideal Gas: Internal energy \(U = f(T)\) (only depends on Temperature).
• Real/van der Waals Gas: Internal energy \(U = f(T, V)\) (both Temperature and Volume matter due to attractive intermolecular forces represented by the parameter \(a\)).

Answer 1

The Correct Option is B

Concept: For an ideal gas, intermolecular forces are assumed to be non-existent, meaning internal energy depends strictly on kinetic energy and is a function of temperature alone (\(U = f(T)\)), as stated by Joule's Law. However, a real gas modeled by the van der Waals equation of state accounts for long-range attractive forces between molecules, introducing a volume dependence into the internal energy equation. Thermodynamic Analysis:
The general mathematical relationship showing how internal energy shifts with volume under isothermal conditions is derived from the fundamental thermodynamic relation: \[ dU = T\,dS - P\,dV \] Dividing by \(dV\) at constant temperature yields: \[ \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P \] Applying a Maxwell relation derived from the Helmholtz free energy function, \(\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V\), we can rewrite the expression as: \[ \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P \] Now let us look at the van der Waals equation of state for one mole of gas: \[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \quad \implies \quad P = \frac{RT}{V - b} - \frac{a}{V^2} \] Differentiating this pressure expression with respect to temperature at constant volume: \[ \left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V - b} \] Substituting this derivative back into our internal energy relation: \[ \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{R}{V - b}\right) - \left(\frac{RT}{V - b} - \frac{a}{V^2}\right) = \frac{a}{V^2} \] Integrating this partial differential equation reveals the explicit formula for internal energy \(U\): \[ U(T, V) = C_v T - \frac{a}{V} + \text{constant} \] This proves that the internal energy \(U\) of a van der Waals gas is a function of both temperature (\(T\)) and volume (\(V\)), which corresponds to Option (2).