Question

A fixed mass of gas ( $γ$ = 1.4) at 1 atm is compressed adiabatically to 5 atm. and then allowed to expand isothermally to its original volume. The pressure at the end of the isothermal process will be

Difficult

Solution

Let P₁,V₁ be the initial pressure and volume and P₂,V₂ those after adiabatic compression respectively. Then
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} or {(V_{2} / V_{1})}^{γ} = (P_{1} / P_{2})$
$∴ \frac{V_{2}}{V_{1}} = {(\frac{P_{1}}{P_{2}})}^{1 / γ} = {(\frac{1}{5})}^{1 / 1 \cdot 4}$
$\begin{array}{l} \log (\frac{V_{2}}{V_{1}}) = \frac{1}{1 \cdot 4} [\log 1 - \log 5] \\ = \frac{1}{1 \cdot 4} [0 - 0 \cdot 6990] \\ = - 0 \cdot 4993 = \bar{1} \cdot 5007 \\ \frac{V_{2}}{V_{1}} = 0 \cdot 3167 \end{array}$
The gas now expands isothermally to original volume V₁ . Let P₃ be the pressure after this expansion. Then
$\begin{array}{l} P_{3} V_{1} = P_{2} V_{2} \\ P_{3} = P_{2} (V_{2} / V_{1}) = (5 atm .) \times 0 \cdot 3167 = 1 \cdot 58 atm \end{array}$

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