Two moles of an ideal monoatomic gas trebles its initial volume in an isobaric expansion from state A to state B. The gas is then cooled isochorically to state C and finally compressed isothermally until it returns to state A. The molar gas constant is R = 8.314 J mol–1 K–1 and Boltzmann's constant is 1.38 × 10–23 J K–1.
State B corresponds to a pressure P = 8 atm (1 atm = 1.013 × 105 Pa) and temperature T = 552°C. (I also know that temperature of the gas in state A is 275 K)
What is the work done on the surroundings in each of the following processes:
The Attempt at a Solution
(i) This process is isobaric
p = constant = pB
I could calculate the volumes from the given information. But it easier to substitute pV = nRT for each state and get:
[tex]W = nR(T_B - T_A)[/tex]
W = 8(1.013 × 105) ∙ (8.314).(825.15 K - 275.05 K)
W= 3706389847 J
Why is the value for work so large? The correct answer to this part must be 9150.
Since we have an isothermal process:
[tex]pV = nRT = constant = nRT_C[/tex]
W = nRTA ∙ln(VA/VC)
W = 8(1.013 × 105) ∙ 8.314 ∙ 275K ∙ ln (VA/VC)
I'm stuck. How do I get the volumes for states A and C?
(P.S. I also know that the work done in the processe B→C is 0, because it is an isochoric (i.e. isovolumetric) process).