10 mol of an ideal gas is taken through an isothermal process in which the volume is compressed from 40 L to 30 L. If the temperature and the pressure of the gas are `0^(@)C` and 1 atm respectively. Find the work done in the process.
Given: `R = 8.31 J cdot mol^(-1) cdot K^(-1)`.

Explanation

Work done for n mol of gas, due to change in volume
in an isothermal process,
`W = nRT In(V_f)/(V_i) = nRT xx 2.3026 log_(10)(V_f)/(V_i)`
`= 10 xx 8.31 xx 273 xx 2.3026 log_(10)30/40`
` = 10 xx 8.31 xx 273 xx 2.3026 log_(10) 3/4`
`= 10 xx 8.31 xx 273 xx 2.3026[0.4771 - 0.6020]`
`= -10 xx 8.31 xx 273 xx 2.3026 xx 0.1251`
`= -6534.91 J`
[the negative sing shows that work is done on the gas].