Answer to Question #215464 in Mechanical Engineering for Aron

Question

Assume that a 2 kg of a certain gas with R=0.4305 kJ/kg-K undergoes a process that results in these changes: ΔU=1458.01 kJ, ΔH=2189.13 kJ. If the process had been internally reversible with p=C, what is the work done?

Answer 1

$C_p-C_v = R$ ...(1)

$k = \frac{C_p}{C_v}$ ...(2)

From 1 and 2

$R = 1.49C_v-C_v = 0.49C_v$

$C_v = \frac{0.4305}{0.49}=0.8785$ $\frac {kJ} {kg K}$

$Q = mC_v\triangle U$

$Q =2 \times 0.8785 \times 1458.01 = 2561.72 kJ$

work done $W = Q - \triangle U =2561.72 - 1458.01 =1103.71 kJ$