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 $3R/2$. [Given $(3{)}^{5/3}=6.19$ ]

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

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]