Deepak Patra
Last Activity: 11 Years ago
Sol. P base 1 = 10 kpa = 10 × 10^3 pa. P base 2 = 50 × 10^3 pa. v base 1 = 200 cc. v base 2 = 50 cc
(i) Work done on the gas = ½ (10+50) x 10^3 x (50-200) x 10 ^-6 = - 4.5 J
(ii) dQ = 0 ⇒ 0 = du + dw ⇒ du = – dw = 4.5 J