Find out the work done in adiabatic compression of 1 mol of an ideal gas. The initial pressure and volume of the gas are `10^(5) N cdot m^(-2)` and 6L, respectively, the final volume is 2L , the molar specific heat of the gas at constant volume is `3/2 R`.

Explanation

Here, `C_(v) = 3/2 R`
As `C_(p) - C_(v) = R, C_(p) = C_(v) + R = 3/2 R + R = 5/2 R`,
so, `gamma = (C_p)/(C_v) = 5/3`
For 1 mol of the gas, `pV = RT or, T = 1/R pV`,
the fall in temperature is `T_(i) - T_(f) = 1/R (p_(i)V_(i) - p_(f)V_(f))`
For this adiabatic process, `p_(i)V_(i)^(gamma) = p_(f)V_(f)^(gamma)`
or, `p_(f) = p_(i) ((V_i)/(V_f))^(gamma) = 10^(5) xx (6/2)^(5//3) = 6.24 xx 10^(5)N cdot m^(-2)`
`:.` Work done ,
`W = R/(gamma - 1) (T_i - T_f) = R/(gamma-1) cdot 1/R (p_(i)V_(i) - p_(f)V_(f))`
`=(p_(i)V_(i) - p_(f)V_(f))/(gamma -1)`
`=(10^(5) xx 6 xx 10^(-3) - 6.24 xx 10^(5) xx 2 xx 10^(-3))/(5/3 - 1)`
`[1 L = 10^(-3) m^(3)]`
`= 3/2 xx (6 xx 10^(2) - 6.24 xx 2 xx 10^(2))`
`=3/2 xx 10^(2) xx (6 - 12.48) = -3/2 xx 10^(2) xx 6.48`
`= -972 J`.