# Homework Help: Work by an ideal gas, thermodynamics

1. Nov 18, 2009

### v0id19

1. The problem statement, all variables and given/known data

At point D: P=2atm, T=360K, n=2mol
At point B: V=3VD, P=2Pc
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.

2. Relevant equations
PV=nRT
W=nRT*ln(V2/V1) (by integration)

3. 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: Pa*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 Wab=-130J and Wcd=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....

2. Nov 18, 2009

### tiny-tim

Hi v0id19!
Perhaps I'm misunderstanding the problem, but isn't work done = ∫PdV, the area inside the graph?

3. Nov 18, 2009

### v0id19

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

4. Nov 18, 2009

### v0id19

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

5. Nov 19, 2009

### tiny-tim

Hi v0id19!

(just got up :zzz: …)
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)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook