Calculate the magnitude of work done (in Joule) when 1 mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are $10^{5} N / m^{2}$ and 6 litres respectively. The final volume of the gas is 2 litre. Molar specific heat of the gas at constant volume is $\frac{3 R}{2}$ . (Take 3^5/3=6.19)

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answer is 957.

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Detailed Solution

$Given that C_{v} = \frac{3}{2} R$

$\Rightarrow γ = \frac{5}{3} (Monoatomic)$

$P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ}$

$∴ P_{2} = 10^{5} {[\frac{6}{2}]}^{\frac{5}{3}} = 10^{5} \times (3)^{\frac{5}{3}} = 6.19 \times 10^{5} N / m^{2}$

$W = \frac{1}{1 - γ} [P_{2} V_{2} - P_{1} V_{1}]$

$∴ W = \frac{1}{1 - \frac{5}{3}} [6.19 \times 10^{5} \times 2 \times 10^{- 3} - 10^{5} \times 6 \times 10^{- 3}] = - 957 J$

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