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Constant Internal Energy Process

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A tank having a volume of 0.1 m^{3} contains air at 14 MPa and 50^{o}C. It is connected through a valve to a larger tank having a volume of 15 m^{3}, which is completely evacuated. The entire assembly is completely insulated. The valve is opened and the gas allowed to come to equilibrium in both tanks. Calculate the final pressure.

Calculations:

For this process, Q = 0 and W = 0. Therefore, DU = 0.

DU = mCvDT = 0.

Therefore, DT = 0.

For an ideal gas, PV = nRT

Since T is constant, P_{1}V_{1} = P_{2}V_{2}

14 x 0.1 = P_{2} x (0.1 + 15)

i.e., final pressure

P_{2} = 92.715 KPa.

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Last Modified on: 11-Sep-2014

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