Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. The initial pressure and…
Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are $10^{5} \mathrm{~N} / \mathrm{m}^{2}$ and 6 litre respectively. The final volume of the gas is 2 litres. Molar specific heat of the gas at constant volume is 3R/2. [Given $(3)^{5 / 3}=6.19$ ]
$-957 \mathrm{~J}$
$+957 \mathrm{~J}$
$-805 \mathrm{~J}$
$+805 \mathrm{~J}$
Solution
For an adiabatic change $\mathrm{PV}^{\gamma}=$ constant $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$
As molar specific heat of gas at constant volume
$\begin{aligned} & C_{v}=\frac{3}{2} R \\ C_{P}=C_{V}+R=\frac{3}{2} R+R=\frac{5}{2} R \\ \gamma=& \frac{C_{P}}{C_{V}}=\frac{(5 / 2) R}{(3 / 2) R}=\frac{5}{3} \\ \therefore \text { From eq }^{n} \cdot(1) \\ & P_{2}=\left(\frac{V_{1}}{V_{2}}ight)^{\gamma} P_{1}=\left(\frac{6}{2}ight)^{5 / 3} \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \\=&(3)^{5 / 3} \times 10^{5}=6.19 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \end{aligned}$
Work done
$=\frac{1}{1-(5 / 3)}\left[6.19 \times 10^{5} \times 2 \times 10^{-3}-10^{-5} \times 6 \times 10^{-3}ight]$
$=-\left[\frac{2 \times 10^{2} \times 3}{2}(6.19-3)ight]$
$=-3 \times 10^{2} \times 3.19=957$ joules
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