Question

For a gas $C_P - C_V = R$ in a state P and $C_P - C_V = 1.10R$ in a state Q, $T_P$ and $T_Q$ are the temperatures in two different states P and Q respectively. Then

Answer 1

$T_P > T_Q$ and $P_P < P_Q$

Answer 2

Explanation of the solution:

1. Analyze State P: For state P, the given condition is $C_P - C_V = R$. This is Mayer's relation, which is strictly valid only for an ideal gas. Therefore, in state P, the gas is behaving as an ideal gas.

2. Analyze State Q: For state Q, the given condition is $C_P - C_V = 1.10R$. Since $1.10R \neq R$, this indicates that the gas in state Q is not behaving ideally; it is exhibiting real gas behavior.

3. Conditions for Ideal and Real Gas Behavior:

   • Gases behave most ideally (adhere to ideal gas laws) at high temperatures and low pressures.
   • Gases deviate from ideal behavior (exhibit real gas properties) at low temperatures and high pressures.

4. Compare States P and Q:

   • Since the gas in state P behaves ideally and in state Q it behaves as a real gas, the conditions in state P must be more conducive to ideal behavior than in state Q.
   • This implies that the temperature in state P must be higher than in state Q ($T_P > T_Q$).
   • Concurrently, the pressure in state P must be lower than in state Q ($P_P < P_Q$).

Answer: The relationship between the temperatures and pressures in the two states is: $T_P > T_Q$ and $P_P < P_Q$