Question
In: Chemistry
in this example, 2.5 mol of an ideal gas with CV,m= 12.47Jmol-1K-1 is expanded adiabatically against...
in this example, 2.5 mol of an ideal gas with CV,m= 12.47Jmol-1K-1 is expanded adiabatically against a constant external pressure of 1 bar. The initial temperature and pressure of the gas are 325 K and 2.5bar respectively. The final pressure is 1.25 bar. Calculate the final temperature, q, w, ΔU and ΔH
Solutions
Expert Solution
Initial volume (V1) = nRT1/P1
= 2.5 x 0.08205 x 325/2.5
= 26.7 L
Final volume (V2) = P1V1/P2
= 2.5 x 26.7/1.25
= 53.4 L
Final Temperature (T2) = T1V2/V1
= 325 x 53.4/26.7
= 650 K
w = nCvdT
= 2.5 x 12.47(650-325
= 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
= 2.5 x 20.784(325)
= 16.90 kJ
queen_honey_blossom answered 1 year agoRelated Solutions
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