For a reversible adiabatic ideal gas expansion dP/P is equal to

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Solution

Q.
For a reversible adiabatic ideal gas expansion $\frac{d P}{P}$ is equal to

1 $γ \frac{d V}{V}$

2 $- γ \frac{d V}{V}$

3 $(\frac{γ}{γ - 1}) \frac{d V}{V}$

4 $\frac{d V}{V}$

Ans.
(2)

$P V^{γ} = K (K is constant)$ $d P V^{γ} + P γ V^{γ - 1} d V = 0$

$\frac{d P}{P} - \frac{γ V^{γ - 1}}{V^{γ}} d V$ $\frac{d P}{P} = - γ \frac{d V}{V}$

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