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4 years ago

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A horizontal piston-cylinder compresses adiabatically 1 mol of propane gas, initially at 350 K, from 0.04 to 0.01 m3. Assume ide

al gas conditions with Cp 74 J(mol-K) (a) For a well-oiled piston-cylin der, what is the final gas temperature? What is the compression work done? (b) For a rusted piston-cylinder, the compression work done is found to be 8000 J. What is the final temperature? What is the entropy gener ation?

1 answer:

german4 years ago

7 0

Answer:

(a) Final gas temperature = 419.09 K

Compression work done = -4418.57 J

(b) The final temperature is 224.91 K

The entropy generation is -40.58 J/mol.K

Explanation:

(a) T2 = T1(V1/V2)^gamma - 1

Cv = Cp - R = 74 - 8.314 = 65.686 J/mol.K

gamma = 1 + R/Cv = 1 + 8.314/65.686 = 1+0.13 = 1.13

T1 = 350 K, V1 = 0.04 m^3, V2 = 0.01 m^3

T2 = 350(0.04/0.01)^1.13-1 = 350(4)^0.13 = 350×1.1974 = 419.09 K

Wc = R/gamma-1(T1 - T2) = 8.314/1.13-1(350 - 419.09) = 8.314/0.13 × -69.09 = -4418.57 J

(b) Wc = 8000 J

Wc = R/gamma-1(T1 - T2)

8000 = 8.314/1.13-1(350 - T2)

8000 = 8.314/0.13(350 - T2)

8000×0.13/8.314 = 350 - T2

350 - T2 = 125.09

T2 = 350 - 125.09 = 224.91 K

∆S = Cv.ln(T2/T1) + R.ln(V2/V1) = 65.686ln(350/224.91) + 8.314ln(0.01/0.04) = -29.05 + (-11.53) = -29.05-11.53 = -40.58 J/mol.K

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