GR0177 #37

mrodruck
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Joined: Sun Sep 23, 2012 11:09 pm

GR0177 #37

Postby mrodruck » Sun Sep 23, 2012 11:15 pm

The solution says that the overall work done is -300kJ. However, I can't figure out why it is negative. Shouldn't the work done by the isothermal process be positive, as W=PiVi*ln(Vi/Vf)? In this case, Vi > Vf which makes the ln function positive, and you know by looking at the graph it does more work than the isobaric process (which should be negative as W=-PdV). Therefore, I'd think it would have to be positive.

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HappyQuark
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Joined: Thu Apr 16, 2009 2:08 am

Re: GR0177 #37

Postby HappyQuark » Tue Sep 25, 2012 2:27 am

mrodruck wrote:The solution says that the overall work done is -300kJ. However, I can't figure out why it is negative. Shouldn't the work done by the isothermal process be positive, as W=PiVi*ln(Vi/Vf)? In this case, Vi > Vf which makes the ln function positive, and you know by looking at the graph it does more work than the isobaric process (which should be negative as W=-PdV). Therefore, I'd think it would have to be positive.


http://www.physicsgrad.com/pgre/0177-37

"We can then determine whether the work is negative or positive by the direction of the process. Recall that a clockwise process represents a heat engine and the work is positive. Alternatively, if the process is counter-clockwise, it's a heat pump and the work is negative. This process is counter-clockwise so we choose (D)."

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Izaac
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Re: GR0177 #37

Postby Izaac » Tue Sep 25, 2012 7:46 am

Besides, on purely geometric terms, you integrate the isobaric process from V=2 to V=5, giving a positive work; and then integrate the isotermal process from V=5 to V=2, giving a negative work (that's high school mathematics: integrating a positive f(X) backward from the X axis, as well as integrating a negative f(X) along the X axis, give a negative result).
Judging from the area under the curves, the magnitude of the isotermal process is greater than that of the isobar process. Hence a negative result.




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