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.

Updated On: Apr 13, 2026
  • \( \dfrac{3}{2}PV \)
  • \(PV\)
  • \( \dfrac{4}{3}PV \)
  • \( \dfrac{PV}{2} \)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Concept: For an adiabatic process \[ PV^\gamma = \text{constant} \] For a monoatomic gas \[ \gamma = \frac{5}{3} \] Internal energy change: \[ \Delta U = nC_v\Delta T \] Using ideal gas relation this becomes \[ \Delta U = \frac{P_2V_2 - P_1V_1}{\gamma-1} \]
Step 1:
Use adiabatic relation. \[ P_1V_1^\gamma = P_2V_2^\gamma \] Given \[ V_2 = 27V \] \[ P V^{5/3} = P_2 (27V)^{5/3} \] \[ P_2 = \frac{P}{27^{5/3}} \] \[ 27^{5/3} = (3^3)^{5/3} = 3^5 \] \[ P_2 = \frac{P}{3^5} \]
Step 2:
Calculate change in internal energy. \[ \Delta U = \frac{P_2V_2 - P_1V_1}{\gamma-1} \] \[ = \frac{\left(\frac{P}{3^5}\right)(27V) - PV}{\frac{5}{3}-1} \]
Step 3:
Simplify the expression. \[ = \frac{\frac{P\cdot27V}{3^5} - PV}{\frac{2}{3}} \] \[ = \frac{\frac{PV}{9} - PV}{\frac{2}{3}} \] \[ = \frac{-\frac{8}{9}PV}{\frac{2}{3}} \] \[ \Delta U = -\frac{4}{3}PV \] Since magnitude is required, \[ \boxed{|\Delta U| = \frac{4}{3}PV} \]
Was this answer helpful?
0
0