(i) w = –Pext × ΔV
As expansion taks place into the evacuated bulb, i.e.,
against vacuum, Pext = 0. Henc, w = 0.
For adiabatic process, q = 0
∴ ΔU = q + w = 0 + 0 = 0.
(ii) V = V2 – V1 = 5 – 1 = 4 litres
P = 1 atm ∴ w = – PV
= –1 × 4 litre atm = – 4 litres atm
= – 4 × 101.3 J = – 405.2 J (1 L – atm = 101.3J)
The negative sign implies that the work is done by the system.
For adiabatic process, ΔU = q + w = 0 – 405.2 J = – 405.2 J.