in this example, 2.5 mol of an ideal gas with CV,m= 12.47Jmol-1K-1 is expanded adiabatically against...

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in this example, 2.5 mol of an ideal gas with CV,m=12.47Jmol-1K-1 is expanded adiabatically against a constantexternal pressure of 1 bar. The initial temperature and pressure ofthe gas are 325 K and 2.5bar respectively. The final pressure is1.25 bar. Calculate the final temperature, q, w, Î”U and Î”H

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3.8 Ratings (747 Votes)

Initial volume (V1) = nRT1/P1

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.5 x 0.08205 x 325/2.5

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 26.7 L

Final volume (V2) = P1V1/P2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.5 x 26.7/1.25

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 53.4 L

Final Temperature (T2) = T1V2/V1

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 325 x 53.4/26.7

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 650 K

w = nCvdT

Â Â Â = 2.5 x 12.47(650-325

Â Â Â = 10.13 kJ

q = 0

dU = w = 10.13 kJ

Cp = Cv + R = 12.47 + 8.314 = 20.784 J.mol-1.K-1

dH = nCpdT

Â Â Â Â Â = 2.5 x 20.784(325)

Â Â Â Â Â = 16.90 kJ

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in this example, 2.5 mol of an ideal gas with CV,m= 12.47Jmol-1K-1 is expanded adiabatically against...

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Chemistry

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