Question

There are 2.5 kg. of gas for which R = 296.86 BTU/kg⁰K and k = 1.399 that undergo a nonflow constant volume process from P1 = 650 kpa and T1 = 65⁰C to P2 = 1745 kpa. During the process, the gas is internally stirred and there are also added 105 KJ of heat.

Answer 1

If we were to calculate the work input

$T_2= \frac{P_2}{P_1}*T_1= \frac{1745}{650}*338=907.4 K= 634.4 ^0 C\\ C_p-C_v= R = 0.313 \\ \frac{C_p}{C_v}=k= 1.399 \implies C_p = 1.399C_v\\ \therefore C_v= 0.7845 kJ/kg ; C_p = 1.0975 kJ/kg\\ dQ=dU+dW\\ 105 = mC_v(T_2-T_1)+dW\\ dW= [0.7845*2.5*(634.4-65)]-105\\ dW= 1011.74 kJ$