Navjyot Kalra
Last Activity: 10 Years ago
Sol. V base 1 = 100 cm^3, V base 2 = 200 cm^3 P = 2 × 10^5 Pa, ∆Q = 50J
(a) ∆Q = du + dw ⇒ 50 = du + 20× 10^5(200 – 100 × 10^–6) ⇒ 50 = du + 20 ⇒ du = 30 J
(b) 30 = n × 3/2 * 8.3 * 300 [U = 3/2 nRT for monoatomic]
⇒ n = 2/3*83 = 2/249 = 0.008
(c) du = nC base vdT ⇒ C base v = dndTU/ = 30/0.008 * 300 = 12.5
C base p C base v + R = 12.5 + 8.3 = 20.3
(d) C base v = 12.5 (Proved above)