Deepak Patra
Last Activity: 11 Years ago
Sol. P base 1 = 100 KPa = 10^5 Pa, V base 1 = 400 cm^3 = 400 * 10^-6 m^3, T base 1 = 300 k,
γ = C base P/C base V = 1.5
(a) Suddenly compressed to V base 2 = 100 cm^3
P base 1 V base 1^γ = P base 2 V base 2^γ ⇒ 10^5 (400)^1.5 = P base 2 * (100)^1.5
⇒ P base 2 = 10^5 * (4)^1.5 = 800 KPa
(b) T base 1V base 1^γ-1 = T base 2V base 2^ γ-1 ⇒ 300 * (400)^1.5 – 1 = T base 2 * (100)^1.5-1 ⇒ T base 2 = 300*20/20 = 600 K
(b) Even if the container is slowly compressed the walls are adiabatic so heat transferred is 0.
Thus the values remain, P base 2 = 800 KPa, T base 2 = 600 K.