γ= 1.5, T = 300 k, V = 1Lv =1/2l
(a) The process is adiabatic as it is sudden,
(c) Internal Energy,
dQ = 0, => du = – dw = –(–82.8)J = 82.8 J ≈ 82 J.
(d) Final Temp = 300√2 = 300 × 1.414 × 100 = 424.2 k ≈ 424 k.
(e) The pressure is kept constant.
The process is isobaric.
Work done = nRdT =1/3R× R × (300 – 300√2 ) Final Temp = 300 K