Answer to Question #215464 in Mechanical Engineering for Aron

Question #215464
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?
1
Expert's answer
2021-07-12T03:37:08-0400

CpCv=RC_p-C_v = R ...(1)


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


From 1 and 2

R=1.49CvCv=0.49CvR = 1.49C_v-C_v = 0.49C_v


Cv=0.43050.49=0.8785C_v = \frac{0.4305}{0.49}=0.8785 kJkgK\frac {kJ} {kg K}


Q=mCvUQ = mC_v\triangle U


Q=2×0.8785×1458.01=2561.72kJQ =2 \times 0.8785 \times 1458.01 = 2561.72 kJ


work done W=QU=2561.721458.01=1103.71kJW = Q - \triangle U =2561.72 - 1458.01 =1103.71 kJ



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