Question

An ideal gas at pressure 2.5 × 10⁵ Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure (b) the final temperature and (c) the work done by the gas in the process. Take γ = 1.5

Answer 1

Initial pressure of the gas, P₁ = 2.5 × 10⁵ Pa
Initial temperature, T₁ = 300 K
Initial volume, V₁ = 100 cc

(a) For an adiabatic process,
P₁V₁^γ = P₂V₂^γ
⇒ 2.5 × 10⁵ × V^1.5 = ${\left(\frac{\mathrm{V}}{2}\right)}^{1.5}$ × P₂
⇒ P₂ = 7.07 × 10⁵
= 7.1 × 10⁵ Pa

(b) Also, for an adiabatic process,
T₁V₁^γ−1 = T₂V₂^γ−1
⇒ 300 × (100)^1.5−1 = T₂ × ${\left(\frac{100}{2}\right)}^{1.5-1}$
= T₂ × (50)^1.5−1
⇒ 300 × 10 = T₂ × 7.07
⇒ T ₂ = 424.32 K = 424 K

(c) Work done by the gas in the process,
$$\begin{gathered} W\mathit{=}\frac{nR}{\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{}[T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}] \\ \mathit{=}\frac{P_{\mathit{1}}{V}_{\mathit{1}}}{T\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{}\mathit{}[T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}] \\ =\frac{2.5\times 10}{300\times 0.5}\times (-124) \\ =-20.67\approx -21\mathrm{J} \end{gathered}$$