question_answer

Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are \[{{10}^{5}}N/{{m}^{2}}\] and 6 liter respectively. The final volume of the gas is 2 liters. Molar specific heat of the gas at constant volume is 3R/2. [Given \[{{(3)}^{5/3}}=\text{ }6.19\]]

A) -957 J  

B)        +957 J  

C) -805 J  

D)        + 805 J.

Correct Answer: A

Solution :

[a] For an adiabatic change \[P{{V}^{\gamma }}=constant\] \[{{P}_{1}}V_{1}^{\gamma }={{P}_{2}}V_{2}^{\gamma }\] As molar specific heat of gas at constant volume \[{{C}_{v}}=\frac{3}{2}R\] \[{{C}_{p}}={{C}_{V}}+R=\frac{3}{2}R+R=\frac{5}{2}R;\] \[\gamma =\frac{{{C}_{p}}}{{{C}_{V}}}=\frac{\left( 5/2 \right)R}{\left( 3/2 \right)R}=\frac{5}{3}\] \[\therefore \]From \[e{{q}^{n}}.\left( 1 \right)\] \[{{P}_{2}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}{{P}_{1}}={{\left( \frac{6}{2} \right)}^{5/3}}\times {{10}^{5}}N/{{m}^{2}}\]   \[={{(3)}^{5/.3}}\times {{10}^{5}}=6.19\times {{10}^{5}}N/{{m}^{2}}\] Work done \[=\frac{1}{1-\left( 5/3 \right)}\]             \[\left[ 6.19\times {{10}^{5}}\times 2\times {{10}^{-3}}-{{10}^{-5}}6\times {{10}^{-3}} \right]\]   \[=-3\times {{10}^{2}}\times 3.19=-957\text{ joules}\] [-ve sign shows external work done on the gas]