Consider a sample of 2.75 moles of argon contained in a 1.50 L piston at 298 K.
ID: 526186 • Letter: C
Question
Explanation / Answer
a. work done (w) = nRTln(V2/V1)
= 2.75 x 8.314 x 298 ln(5.05/1.5) = 8270.85 J
q = w = 8270.85 J
dU = 0
dH = 0
b. work done (w) = -nRTln(V2/V1)
= -2.75 x 8.314 x 298 ln(5.05/1.5) = -8270.85 J
q = -w = -8270.85 J
dU = 0
dH = 0
c. work done (w) = nPdV
= 2.75 x (2.75 x 0.08205 x 298/1.5) (5.05 - 1.5) x 101.33 = 44.344 kJ
q = 0
dU = w = 44.344 kJ
Cp = Cv + R = 20.786 + 8.314 = 29.1 J/K
T2 = 1.5 x 298/5.05 = 88.515 K
dH = CpdT = 29.1 (88.515 - 298) = -6096.01 J