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 $6$ litre respectively. The final volume of the gas is $2$ litres. Molar specific heat of the gas at constant volume is $3 R / 2$. [Given $(3)^{5 / 3}=6.19$ ]

• A. $-957\, J$
• B. $+957 \, J$
• C. $-805\, J$
• D. $+805 \, J$

Answer 1

Answer: (A)

For an adiabatic change $PV \gamma=$ 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)
$P _2=\left(\frac{ V _1}{ V _2}\right)^\gamma P _1=\left(\frac{6}{2}\right)^{5 / 3} \times 10^5\, N / m ^2$
$=(3)^{5 / 3} \times 10^5=6.19 \times 10^5 \,N / m ^2$
Work done
$=\frac{1}{1-(5 / 3)}\left[6.19 \times 10^5 \times 2 \times 10^{-3}-10^{-5} \times 6 \times 10^{-3}\right]$
$=-\left[\frac{2 \times 10^2 \times 3}{2}(6.19-3)\right] $
$=-3 \times 10^2 \times 3.19=-957 \text { joules }$
[-ve sign shows external work done on the gas]