Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are `105 N//m^2` and 6 litres respectively. The final volume of the gas are 2 litre. Molar specific heat of the gas at constant volume is `3R//2`.

Answer

`-972`

Explanation

In adiabatic process
`P_1 V_1^gamma = P fV_f^gamma`
` pf=(V_i/V_f)^(gamma) "Pi ...(i)"`
Further, `C_V = (3R)/2 " ":.c_p =C_V +R = (5R)/2`
or `gamma = c_p /c_V = 5/3 " ":." "Pf =(6/2)^(5//3) (10)^5 = 6.24 xx10^5 N//m^2`
Now, work done in adiabatic process is given by
`W = (P_i V_i - PfVf)/(gamma -1)`
`=(10^5 xx6 xx10^(-3)-6.24 xx10^5 xx2xx10^(-3))/((5/3)-1)=-972 J`
Note: - Work done is negative because volume of the gas is decreasing.