Question

Three moles of a gas at a temperature \( T \) are heated to thrice its volume by keeping the pressure constant. If \( \gamma \) is the ratio of specific heats, then the increase in internal energy of the gas is:

• A. \( \frac{3RT}{\gamma - 1} \)
• B. \( \frac{6RT}{\gamma - 1} \)
• C. \( \frac{8RT}{\gamma - 1} \)
• D. \( \frac{3R}{2(\gamma - 1)} \)

Hint:

For a gas undergoing a process at constant pressure, the increase in internal energy can be calculated using the relation \( \Delta U = n C_v \Delta T \), where \( C_v \) is related to \( \gamma \) as \( C_v = \frac{R}{\gamma - 1} \).

Answer 1

The Correct Option is B

Step 1: Use the formula for heat added at constant pressure. \[ Q = n C_p \Delta T \] where \( n = 3 \) moles, \( C_p = \gamma C_v \), and \( \Delta T = 2T \).
Step 2: Use the relation between \( C_p \) and \( C_v \). \[ \Delta U = n C_v \Delta T = n \left( \frac{R}{\gamma - 1} \right) (2T) \]
Step 3: Calculate the increase in internal energy. The increase in internal energy is: \[ \Delta U = 3 \left( \frac{R}{\gamma - 1} \right) (2T) = \frac{6RT}{\gamma - 1} \]