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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

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Solution

## 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, P1V1γ = P2V2γ ⇒ 2.5 × 105 × V1.5 = ${\left(\frac{\mathrm{V}}{2}\right)}^{1.5}$ × P2 ⇒ P2 = 7.07 × 105 = 7.1 × 105 Pa (b) Also, for an adiabatic process, T1V1γ−1 = T2V2γ−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, $W\mathit{=}\frac{nR}{\mathit{\left(}\gamma \mathit{-}\mathit{1}\mathit{\right)}}\mathit{}\left[{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}\right]\phantom{\rule{0ex}{0ex}}\mathit{=}\frac{{P}_{\mathit{1}}{V}_{\mathit{1}}}{T\mathit{\left(}\gamma \mathit{-}\mathit{1}\mathit{\right)}}\mathit{}\mathit{}\left[{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{2.5×10}{300×0.5}×\left(-124\right)\phantom{\rule{0ex}{0ex}}=-20.67\approx -21\mathrm{J}$

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