Find out the heat absorbed and work done when 5 L ofan ideal gas at 10 atm pressure expands to 20L isothermally (1) into vacuum (ii) against a constant external pressure of 1 atm (iii) to a final volume 20L reversibly.

Explanation

(i) Heat,` q = w = P _(ex) (20-5) = 0 (15) = 0 ` Word is done = 0 , Heat absorbed =0.
(ii) `q = w = P _(ex) (V _(f) - V _(i)) = 2 atm xx (20 - 5) L = 30 L atm.`
(iii)` q =-w 2.303 RT log (V _(2) // V _(1)) = 2.303 PV log (V _(2) // V _(1)) = 2.303 PV log (V _(2)) // V _(1))`
`q = 2.303 xx 10 atm xx 5 L xx log (20 //5) = 69.33 L atm = 69.33 xx 101.3 = 7022.85J`