Question

Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an ideal gas at P = 100kPa and T = 500K; on the right is a vacuum. Now the wall is removed. a) What is the final temperature in the tank? b) What is the ∆Suniv due to the removal of the wall?

Answer 1

given,

V = 2 L

the left is an ideal gas at P = 100 k Pa and T = 500 K

mass is constant

`m_1 = m_2`

`(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2)`

Pressure is same because it's not changing due to process

`(V)/(500) = (2 V)/(T_2)`

`T_2 = 1000\ K`

`\Delta S_(univ) = \Delta S_(sys) + (\Delta S)_(surr)`

`\Delta S_(univ) =m(C_v ln ((T_2)/(T_1)))+ R ln ((V_2)/(V_1))`

`m = (P_1V_1)/(RT_1)`

`m = (100 * 10^3 * 2 * 10^(-3))/(287* 500)`

m = 1.39 x 10⁻³ Kg

`\Delta S_(univ) =1.39* 10^(-3)(0.718 ln\ 2+ 0.287 ln (2)`

`\Delta S_(univ) =0.968* 10^(-3)\ kJ/K`