Work by an ideal gas, thermodynamics

[v0id19]

Homework Statement

At point D: P=2atm, T=360K, n=2mol
At point B: V=3V_D, P=2P_c
Paths AB and CD represent isothermal processes
The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas.

Homework Equations

PV=nRT
W=nRT*ln(V₂/V₁) (by integration)

The Attempt at a Solution

I only need the work for the two curved sections (because the work by the vertical ones is zero). Using a system of equations, I determined the following (in the form PV=nRT):
A: P_a*29.6=2*R*721
B: (4/3)*88.8=2*R*721
C: (2/3)*88.8=2*R*360
D: 2*29.6=2*R*360

Using the work equation, I get W_ab=-130J and W_cd=64.9J

But the solutions in my book say I should have -13.2kJ and 6.58kJ, respctively


I am pretty sure I solved for the variables correctly; am I not using the correct formula to find work? Or is there something I've missed...

 

[tiny-tim]

Hi v0id19!

PV=nRT
W=nRT*ln(V₂/V₁) (by integration)

Perhaps I'm misunderstanding the problem, but isn't work done = ∫PdV, the area inside the graph?

 

[v0id19]

yes, but I don't have any of the equations for the lines on the graph.
using the energy equation (E=Q_in+W). for an isothermal process E=0, so Q_in=-W, and W=PdV=∫(nRT)/V*dV=nRT*ln(V₂/V₁)

 

[v0id19]

actually--i just redid the problem, it turns out i was using the wrong gas constant R...

 

[tiny-tim]

Hi v0id19!

(just got up :zzz: …)

yes, but I don't have any of the equations for the lines on the graph.

But you are given the width and height of the region as proportions of the (x,y) coordinates of D.

(and, assuming AB and DC are meant to be "parallel", the area is width times height)