- #1

roam

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## Homework Statement

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 × 10

^{5}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:

**(i)**A→B

**(ii)**C→A

## The Attempt at a Solution

**(i)**This process is isobaric

p = constant = p

_{B}

[tex]W=-P(V_B-V_A)[/tex]

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 × 10

^{5}) ∙ (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**.

**(ii)**

Since we have an isothermal process:

[tex]pV = nRT = constant = nRT_C[/tex]

W = nRT

_{A}∙ln(V

_{A}/V

_{C})

W = 8(1.013 × 10

^{5}) ∙ 8.314 ∙ 275K ∙ ln (V

_{A}/V

_{C})

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).