Question

Initial pressure and volume of a monoatomic ideal gas are \(P\) and \(V\). The change in internal energy of this gas in adiabatic expansion to volume \(V_{\text{final}} = 8V\) is ________ J.

• A. \( -2PV(3\sqrt{3} - 1) \)
• B. \( \frac{4}{3}PV \)
• C. \( -\frac{3}{4}PV \)
• D. \( \frac{3}{4}PV \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an adiabatic process, there is no heat exchange ($Q=0$). The change in internal energy equals the negative of the work done by the gas, or can be calculated directly using the final and initial states of pressure and volume.
Step 2: Key Formula or Approach:
1. Adiabatic equation: $P_1 V_1^\gamma = P_2 V_2^\gamma$.
2. Monoatomic gas specific heat ratio: $\gamma = \frac{5}{3}$.
3. Change in internal energy: $\Delta U = \frac{f}{2}(P_2 V_2 - P_1 V_1)$ where $f=3$ for a monoatomic gas.
Step 3: Detailed Explanation:
Given:
Initial State: $P_1 = P, V_1 = V$.
Final State: $V_2 = 27V$.
Using the adiabatic condition to find $P_2$:
$P_1 V_1^{5/3} = P_2 V_2^{5/3}$.
$P (V)^{5/3} = P_2 (27V)^{5/3}$.
$P = P_2 (27)^{5/3} = P_2 (3^3)^{5/3} = P_2 (3^5) = 243 P_2$.
Therefore, $P_2 = \frac{P}{243}$.
Now, calculate the change in internal energy $\Delta U$. For a monoatomic gas, degrees of freedom $f=3$.
$\Delta U = n C_v (T_2 - T_1) = \frac{3}{2} n R (T_2 - T_1) = \frac{3}{2} (P_2 V_2 - P_1 V_1)$.
Substitute the values of pressures and volumes:
$\Delta U = \frac{3}{2} \left[ \left( \frac{P}{243} \right) (27V) - PV \right]$.
$\Delta U = \frac{3}{2} \left[ \frac{27}{243} PV - PV \right]$.
Since $\frac{27}{243} = \frac{1}{9}$:
$\Delta U = \frac{3}{2} \left[ \frac{1}{9} PV - PV \right] = \frac{3}{2} \left[ -\frac{8}{9} PV \right]$.
$\Delta U = -\frac{24}{18} PV = -\frac{4}{3} PV$.
Step 4: Final Answer:
The change in internal energy is $-\frac{4}{3}PV$.