GR9677 #73



Alternate Solutions 
griffitz 20121107 18:57:42  The process is adiabatic, so dQ= 0. Thus, dW = dU, or dW = f/2 * NkdT. But since T = PV/Nk, we have:
dW = f/2 * d(PV).
d(PV) =
and = (1+2f)/f, thus
= f\2
and we arrive at choice (C)  

Comments 
griffitz 20121107 18:57:42  The process is adiabatic, so dQ= 0. Thus, dW = dU, or dW = f/2 * NkdT. But since T = PV/Nk, we have:
dW = f/2 * d(PV).
d(PV) =
and = (1+2f)/f, thus
= f\2
and we arrive at choice (C)   timtamm 20110825 21:00:37  i feel like there is a problem with the power in the solution to the integral it should be 1gamma not gamma 1 in the power of V. am i crazy or is this a typo someone else back me up
timtamm 20110825 21:01:59 
maybe i am just forgetting that it is in the demoninator still god this math is driving me bananas  i'm thinking just skipping these problems on the test

arboretum 20120702 13:54:47 
if you look at the problem, it asks what is the work done BY the gas. In other words, work done ON the surroundings. The expression for work is therefore going to be opposite of what it usually is (positive instead of negative).

arboretum 20120702 13:58:17 
oh never mind, you were asking about the power, not the coefficient.Oops.

  olj5 20110803 08:44:15  What is gamma=1?
olj5 20110803 08:45:31 
What IF* is what I mean.

pam d 20110917 21:05:58 
Gamma is the ratio of Cp to Cv, and as far as I know they are never equal. Normally the relation is Cp = Cv +R where R is the familiar gas constant. Cp is the molar heat capacity at constant pressure and Cv is the molar heat capacity at constant volume.

  archard 20100530 13:30:19  Once you see that you have an integrand , you can pretty much bet that the answer is C since it's the only one with a in the denominator.   Inri 20100409 23:29:04  You can immediately eliminate (D) and (E) by knowing that the numerator can't depend that way on gamma, as it produce the wrong units for anything but gamma = 0. You can immediately eliminate (A) by knowing that the work depends on the initial pressure and volume.
You can eliminate (B) by knowing that the sum of the pressures shouldn't affect the final work (though the difference might).
That leaves option (C).
Fortunately, the integral is not too difficult and in this problem you only need to get so far as the constant or as there is only one option with that constant.
flyboy621 20101105 20:31:53 
well said

  a19grey2 20081030 21:07:21  If we plug in C's dependence on V later in the problem, why can we ignore this dependence when doing the integral?
gt2009 20090703 12:35:10 
C is a constant so you can pull it out of the integral. It is not ignoring anything.

  physicsisgod 20081030 15:07:48  It took me a minute to realize how , so in case anyone else has this problem:
Tada! Thanks, ETS, for putting the correct answer in the most laborious form possible!
AER 20090331 17:15:29 
Thankfully, once you get the factor, you're done.

sullx 20091103 18:54:33 
Yeah. After I saw the factor I was confident of C.

 




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