Q101 / 125 · BITSAT 2012 · Physics

For an isothermal expansion of a perfect gas, the value of $\frac{\Delta P}{P}$ is equal to

  • A
    $-\gamma^{1 / 2} \frac{\Delta V}{V}$
  • Correct answer
    B
    $-\frac{\Delta V}{V}$
  • C
    $-\gamma \frac{\Delta V}{V}$
  • D
    $-\gamma^{2} \frac{\Delta V}{V}$

Solution

Differentiate $P V=$ constant w.r.t $V$ $\Rightarrow P \Delta V+V \Delta P=0 \quad \Rightarrow \frac{\Delta P}{P}=-\frac{\Delta V}{V}$

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

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