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.
(1) Ratio of the two specific heats for an ideal monoatomic gas = = 1.67.
Initial pressure = Pi = 1.4 x 105 Pa, and, initial volume = Vi = 0.5 m3.
Hence, PiVi = 1.4 x 105 x 0.51.67 = 4.4 x 104 = K, say.
Final volume of the gas = Vf = 1.75 m3.
Hence, adiabatic work done = W = K ( Vf1 - - Vi1 - ) / ( 1 - )
or, W = 4.4 x 104 [ ( 1.751 - 1.67 - 0.51 - 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.
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