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Thermodynamics

Question

For an adiabatic expansion of a monoatomic perfect gas, the volume increases by 24%. What is the percentage decrease in pressure?

24%

40%

48%

71%

Solution

40%

As we know that according to adiabatic expansion

pV^γ = K

γ is the ratio of specific heat.

By differentiating

∴ Δp V^γ + pγ V^γ-1 ΔV = 0

⇒ $(\frac{∆ p}{p}) = - γ (\frac{∆ V}{V})$
For monoatomc gas,

$γ = \frac{5}{3}$

$\frac{∆ p}{p} = \frac{5}{3} \times 24 %$

$\frac{∆ p}{p}$ = 40 %