Three moles of helium gas at one atmosphere are compressed reversibly and isothermally at `127^(@)C` to 5 atmospheres. Calculate the work done.

Explanation

`T = 127^(@)C = 127 + 273 = 400 K, P_(1) = 1` atm, `P_(2) = 5` atm
`R = 8.314 J//K mol, n = 3 "mole"`
`w = -2.303 nRT log ((P_(1))/(P_(2))) = -2.303 xx 3 xx 8.314 xx 400 xx log ((1)/(5))`
`= -2.303 xx 3 xx 8.134 xx 400 xx bar(1).3010`
`= -2.303 xx 3 xx 8.314 xx 400 xx (-0.6990) = +16,060 "joule" = +16.06 kJ`