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GR9677 #73
Problem
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Thermodynamics$\Rightarrow$}Adiabatic Work

One should recall the expression for work done by an ideal gas in an adiabatic process. But, if not, one can easily derive it from the condition given in the problem, viz., $PV^\gamma =C \Rightarrow P=C/V^\gamma$.

Recall that the definition of work is $W=\int PdV =\int_{V_1}^{V_2} C dV/V^\gamma =\left.-\frac{1}{\gamma-1} C/V^{\gamma -1} \right|_{V_1}^{V_2}$, which when one plugs in the endpoint limits, becomes choice (C).

Alternate Solutions
 griffitz2012-11-07 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) = $P_f *V_f - P_i*V_i$ and $\gamma$ = (1+2f)/f, thus $\frac{1}{1-\gamma}$ = f\2 and we arrive at choice (C)Reply to this comment
griffitz
2012-11-07 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) = $P_f *V_f - P_i*V_i$

and $\gamma$ = (1+2f)/f, thus

$\frac{1}{1-\gamma}$ = f\2

and we arrive at choice (C)
timtamm
2011-08-25 21:00:37
i feel like there is a problem with the power in the solution to the integral it should be 1-gamma not gamma -1 in the power of V. am i crazy or is this a typo--- someone else back me up
 timtamm2011-08-25 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
 arboretum2012-07-02 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).
 arboretum2012-07-02 13:58:17 oh never mind, you were asking about the $\gamma-1$ power, not the coefficient.Oops.
olj5
2011-08-03 08:44:15
What is gamma=1?
 olj52011-08-03 08:45:31 What IF* is what I mean.
 pam d2011-09-17 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
2010-05-30 13:30:19
Once you see that you have an integrand $\frac{dV}{V^\gamma}$, you can pretty much bet that the answer is C since it's the only one with a $\1-\gamma$ in the denominator.
Inri
2010-04-09 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 $(\gamma - 1)$ or $(1 - \gamma)$ as there is only one option with that constant.
 flyboy6212010-11-05 20:31:53 well said
a19grey2
2008-10-30 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?
 gt20092009-07-03 12:35:10 C is a constant so you can pull it out of the integral. It is not ignoring anything.
physicsisgod
2008-10-30 15:07:48
It took me a minute to realize how $\frac{C}{V^{\gamma -1}} = PV$, so in case anyone else has this problem:

$\frac{C}{V^{\gamma -1}} = \frac{C}{V^\gamma V^{-1}} = \frac{C}{V^\gamma} V= PV$

Tada! Thanks, ETS, for putting the correct answer in the most laborious form possible!
 AER2009-03-31 17:15:29 Thankfully, once you get the $1-\gamma$ factor, you're done.
 sullx2009-11-03 18:54:33 Yeah. After I saw the $1-\gamma$ factor I was confident of C.

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