Question #215465

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. What is the value of cv?

Expert's answer

△H=CP△T\triangle H= C_P\triangle T


△U=CV△T\triangle U= C_V\triangle T


Specific heat ratio, k=CpCv=△H△U=2189.131458.01=1.49k = \frac{C_p}{C_v} =\frac {\triangle H}{\triangle U}= \frac {2189.13}{1458.01}=1.49


Cp−Cv=RC_p-C_v = R ​...(1)


k=CpCvk = \frac{C_p}{C_v} ​...(2)


From 1 and 2

R=1.49Cv−Cv=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}

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