Q117 / 155 · WBJEE 2022 · Physics

Consider a thermodynamic process where internal energy $U=A P^2 \mathrm{~V}(\mathrm{~A}=$ constant). If the process is performed adiabatically, then

  • A
    $A P^2(V+1)=$ constant
  • Correct answer
    B
    $(\mathrm{AP}+1)^2 \mathrm{~V}=\mathrm{constant}$
  • C
    $(\mathrm{AP}+1) \mathrm{V}^2=$ constant
  • D
    $\frac{V}{(A P+1)^2}=$ constant

Solution

$U=A P^2 V$ $\frac{d U}{d V}=A P^2+2 P A V \frac{d P}{d V}$ [For adiabatic process $\left.d Q=0, d U=-d w=-P d V\right]$ $-\frac{P d V}{d V}=A P^2+2 P A V \frac{d P}{d V}$ $-1=A P+2 A V \frac{d P}{d V} \Rightarrow-1-A P=2 A V \frac{d P}{d V}$ $-\int \frac{d V}{V}=\int \frac{2 A d P}{A P+1}$ $-\ell n V+C=\frac{2 A}{A} \ln (A P+1)$ $C=\ell n(A P+1)^2+\ell n V \Rightarrow C=(A P+1)^2 V$

Hence, the correct answer is option: 'B'.

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