Thermodynamics Question 268
Question: 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}}=6.19 $ ]
Options:
A) -957 J
B) +957 J
C) -805 J
D) + 805 J.
Show Answer
Answer:
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{( 5/2 )R}{( 3/2 )R}=\frac{5}{3} $
$ \therefore $
From $ eq^{n}.( 1 ) $
$ P _{2}={{( \frac{V _{1}}{V _{2}} )}^{\gamma }}P _{1}={{( \frac{6}{2} )}^{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-( 5/3 )} $
$ [ 6.19\times 10^{5}\times 2\times {{10}^{-3}}-{{10}^{-5}}6\times {{10}^{-3}} ] $
$ =-3\times 10^{2}\times 3.19=-957\text{ joules} $ [-ve sign shows external work done on the gas]
