JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
For a gas C P - C V = R in a state P and C P - C V = 1.10 R in a state Q, T P and T Q are the temperatures in two different states P and Q respectively. Then
- (a) T P = T Q
- (b) T P < T Q
- (c) T P = 0.9 T Q
- (d) T P > T Q
Answer and solution
Answer (d) T P > T Q
Solution
Physics sir · 4 steps
C_P - C_V = R is the ideal-gas mark, and a gas gets closer to ideal as it gets hotter.
- Ideal gas benchmark. Mayer's relation C_P - C_V = R holds exactly only for an ideal gas, where molecules do not attract each other. In state P the gas meets it exactly, so it behaves ideally there.
- Real gas deviation. In state Q the difference is 1.10R, so the gas is acting as a real gas. Intermolecular attraction makes C_P - C_V larger than R. For a van der Waals gas at low density, this works out to:
C_P - C_V ≈ R(1 + 2a/RTV)
- Role of temperature. The extra term shrinks as T rises, because fast molecules hardly feel the attraction. So a larger deviation from R means a lower temperature.
- So. State P, which is ideal, must be hotter than state Q, which deviates. So T_P > T_Q, option (d).
The idea to keep. Real gases behave ideally at high temperature and low pressure. Any deviation from an ideal-gas result, such as C_P - C_V = R, points to a colder or more compressed state.
Answer from the NTA final answer key. Solution by PrepWiser.