Question

The change in internal energy of given mass of a gas, when its volume changes from \( V \) to \( 3V \) at constant pressure \( P \) is
(\( \gamma \) - Ratio of the specific heat capacities of the gas)

• A. \( \dfrac{PV}{\gamma - 1} \)
• B. \( \dfrac{2PV}{\gamma - 1} \)
• C. \( \dfrac{3PV}{\gamma - 1} \)
• D. \( \dfrac{PV}{2\gamma - 1} \)

Hint:

In processes at constant pressure, always use the first law of thermodynamics and connect work done and change in temperature using the ideal gas law.

Answer 1

The Correct Option is B

Step 1: Use the First Law of Thermodynamics:
\[ \Delta Q = \Delta U + W \] For a process at constant pressure: \[ \Delta Q = nC_p \Delta T,
\Delta U = nC_v \Delta T \] Step 2: Given volume changes from \( V \) to \( 3V \) at constant pressure \( P \). So work done is: \[ W = P(V_f - V_i) = P(3V - V) = 2PV \] Step 3: Since \( \Delta Q = \Delta U + W \), we rearrange: \[ \Delta U = \Delta Q - W = nC_p \Delta T - W \] But \( C_p - C_v = R \Rightarrow C_v = \frac{R}{\gamma - 1} \) Now use ideal gas law: \[ PV = nRT \Rightarrow nR\Delta T = P\Delta V = 2PV \Rightarrow \Delta T = \frac{2PV}{nR} \] So: \[ \Delta U = nC_v \Delta T = n . \frac{R}{\gamma - 1} . \frac{2PV}{nR} = \frac{2PV}{\gamma - 1} \] \[ \boxed{\Delta U = \frac{2PV}{\gamma - 1}} \]