Need Help Solving for Work Done: 60,000 J

[JFS321]

All, I need some direction on the following question, please.
Correct answer is 60,000 J according to book.

Here's what I have so far, but it may be incorrect: Work done C-->D is (2 atm)(-400 L)(101) = -81,000 J, so work was done on the system when the gas was compressed. No work is done D-->A. Lastly, work done A-->C is (3 atm)(400 L)(101) = 121,200 J. This is work done by the gas on the surroundings. As you can see, my figures do not get to 60,000 J! Thanks in advance. Hopefully I haven't done anything too stupidly obvious!

 

[RUber]

For A-->C, why did you use 3atm? Shouldn't it be the average pressure?

 

[lychette]

units for pressure should be Pa and units for volume should be m³

 

[RUber]

The conversion was handled with the (101) term.
1 L-atm = 101.325 J
Perhaps the problem assumes a simpler conversion factor of approx. 1:100.

 

[JFS321]

Yes, it should be 3.5 atm and this gets me very close to the 60,000 J answer. Thank you.

 

[lychette]

The conversion was handled with the (101) term.
1 L-atm = 101.325 J
Perhaps the problem assumes a simpler conversion factor of approx. 1:100.

The conversion was handled with the (101) term.
1 L-atm = 101.325 J
Perhaps the problem assumes a simpler conversion factor of approx. 1:100.

sorry...I see that the units were taken care of 10³ x 10^-3