Question

If the ratio of specific heat of a gas at constant pressure to that at constant volume is \( \gamma \), the change in internal energy of a mass of a gas when the volume changes from \( V \) to 3\( V \) at constant pressure is:

• A. \( \frac{R}{T ( \gamma - 1)} \)
• B. \( 2 PV \)
• C. \( 2PV \)
• D. \( \frac{2PV}{(\gamma - 1)} \)

Hint:

When dealing with ideal gases and thermodynamics, remember that the change in internal energy for an ideal gas is only dependent on temperature, while the work and heat terms depend on pressure and volume.

Answer 1

The Correct Option is D

The first law of thermodynamics states that: \[ \Delta U = Q - W \] Where: - \( \Delta U \) is the change in internal energy. - \( Q \) is the heat added to the system. - \( W \) is the work done by the system. For an ideal gas undergoing an expansion at constant pressure, the work done is given by: \[ W = P \Delta V = P (V_2 - V_1) \] Now, the heat added at constant pressure is: \[ Q = n C_p \Delta T \] Using the relation for specific heats: \[ C_p = \gamma C_V \] We know the relationship between the specific heat at constant pressure and constant volume. The change in temperature can be calculated from the ideal gas law, where \( V_2 = 3V_1 \), leading to: \[ \Delta U = n C_V \Delta T = \frac{2PV}{(\gamma - 1)} \] Thus, the change in internal energy of the gas is \( \frac{2PV}{(\gamma - 1)} \).