Question

An ideal gas has number of moles \(n=2\), initial volume \(V_0\) and pressure \[ P=P_0\left[1+\left(\frac{V_0}{V}\right)^2\right]^{-1}. \] The gas goes from state \(A\) (initial) to \(B\) (final) such that volume becomes \(3V_0\). Find \(T_A-T_B\).

• A. \( \dfrac{11P_0V_0}{10R} \)
• B. \( \dfrac{5P_0V_0}{11R} \)
• C. \( \dfrac{11P_0V_0}{5R} \)
• D. \( \dfrac{10P_0V_0}{11R} \)

Answer 1

The Correct Option is A

Concept:
For an ideal gas, \[ PV=nRT \] Here \(n=2\). Thus temperature can be found from \[ T=\frac{PV}{2R} \] Step 1: Find temperature at state \(A\). At \(V=V_0\), \[ P=\frac{P_0}{1+(V_0/V_0)^2} =\frac{P_0}{2} \] Using ideal gas equation: \[ \frac{P_0}{2}\times V_0 = 2R\,T_A \] \[ T_A=\frac{P_0V_0}{4R} \] Step 2: Find temperature at state \(B\). At \(V=3V_0\), \[ P=\frac{P_0}{1+\left(\frac{V_0}{3V_0}\right)^2} =\frac{P_0}{1+\frac{1}{9}} =\frac{9P_0}{10} \] Using ideal gas equation: \[ \frac{9P_0}{10}\times3V_0 =2R\,T_B \] \[ T_B=\frac{27P_0V_0}{20R} \] Step 3: Find temperature difference. \[ T_B-T_A= \frac{27P_0V_0}{20R}-\frac{P_0V_0}{4R} \] \[ =\frac{27P_0V_0}{20R}-\frac{5P_0V_0}{20R} \] \[ =\frac{11P_0V_0}{10R} \]