Question

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. $T_P = T_Q$
• D. $T_P = 0.9 T_Q$

Hint:

Ideal gas behavior is approached at high temperature and low pressure.
Any deviation from \(C_p - C_v = R\) suggests non-ideal (real) conditions, typically found at lower temperatures.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Mayer's relation \(C_p - C_v = R\) holds strictly for an ideal gas.
Real gases deviate from this behavior. For real gases, the difference \(C_p - C_v\) is generally greater than \(R\) due to intermolecular forces and molecular size.
Step 2: Key Formula or Approach:
For a real gas (Van der Waals gas), the relation is approximately:
\[ C_p - C_v = R \left( 1 + \frac{2aP}{R^2 T^3} \right) \text{ or } C_p - C_v \approx R + \frac{2a}{vRT} \]
This shows that as temperature \(T\) decreases, the deviation from the ideal value \(R\) increases.
Step 3: Detailed Explanation:
In State P: \(C_p - C_v = R\). This corresponds to ideal behavior, which occurs at high temperatures where intermolecular forces are negligible.
In State Q: \(C_p - C_v = 1.10 R\). This shows a 10% deviation from ideal behavior. Real gases behave more non-ideally at lower temperatures.
Since State P is closer to ideal behavior than State Q, the temperature of State P must be higher than that of State Q.
Therefore, \(T_P>T_Q\).
Step 4: Final Answer:
The relationship is \(T_P>T_Q\).