Question

Initial pressure and volume of monoatomic gas is \(P\) and \(V\). It is expanded adiabatically to \(27\) times of initial volume. Find magnitude of change in internal energy.

• A. \( \dfrac{3}{2}PV \)
• B. \(PV\)
• C. \( \dfrac{4}{3}PV \)
• D. \( \dfrac{PV}{2} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
During an adiabatic expansion, the gas does work at the expense of its internal energy, meaning $Q=0$ and $\Delta U = -W$.
Step 2: Key Formula or Approach:
For an adiabatic process, $PV^\gamma = \text{constant}$.
For a monoatomic gas, $\gamma = 5/3$.
Change in internal energy $\Delta U = n C_V \Delta T = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}$.
Step 3: Detailed Explanation:
Given $V_1 = V$, $V_2 = 27V$, and $P_1 = P$.
Using the relation $P_1 V_1^\gamma = P_2 V_2^\gamma$:
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P \left(\frac{V}{27V}\right)^{5/3} \]
\[ P_2 = P \left(\frac{1}{3^3}\right)^{5/3} = P \left(\frac{1}{3^5}\right) = \frac{P}{243} \]
Now, calculate the change in internal energy $\Delta U$:
\[ \Delta U = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]
\[ \Delta U = \frac{\left(\frac{P}{243}\right)(27V) - PV}{\frac{5}{3} - 1} \]
\[ \Delta U = \frac{\frac{PV}{9} - PV}{2/3} = \frac{\frac{-8PV}{9}}{2/3} \]
\[ \Delta U = -\frac{8PV}{9} \times \frac{3}{2} = -\frac{4}{3}PV \]
The problem asks for the magnitude of change:
\[ |\Delta U| = \frac{4}{3}PV \]
Step 4: Final Answer:
The magnitude of change in internal energy is $\frac{4}{3}PV$.