Question

1. An ideal monatomic gas, with 24.05 moles, expands adiabatically from 0.500m^3 to 1.75 m^3. IF the initial pressure and temperature are 1.40x10^5 Pa and 350K, respectively, find the change in internal energy of the gas if the final temperature of the gas is 152K.

2. A fridge does 17.5 KJ of work while moving 120KJ of thermal energy from inside the fridge. Calculate the fridge's coefficient of performance.

Answer 1

(1) Ratio of the two specific heats for an ideal monoatomic gas = = 1.67.

Initial pressure = P_i = 1.4 x 10⁵ Pa, and, initial volume = V_i = 0.5 m³.

Hence, P_iV_i = 1.4 x 10⁵ x 0.5^1.67 = 4.4 x 10⁴ = K, say.

Final volume of the gas = V_f = 1.75 m³.

Hence, adiabatic work done = W = K ( V_f^{1 -} - V_i^{1 -} ) / ( 1 - )

or, W = 4.4 x 10⁴ [ ( 1.75^{1 - 1.67} - 0.5^{1 - 1.67} ) / ( 1 - 1.67 ) ] J

or, W = 59350 J.

there is no heat exchange in an adiabatic process, so, heat input = Q = 0 J.

Hence, from the first law of thermodynamics,

change in the internal energy = | Q - W | = 59350 J.

(2) Coeffitient of performance of the fridge = Heat taken out / Work done = 120 kJ / 17.5 kJ = 6.875.