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GR9277 #62
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{62}
A mole of ideal gas initially at temperature $T_0$ and volume $V_0$ undergoes a reversible isothermal expansion to volume $V_1$. If the ratio of specific heats is $c_p/c_v=\gamma$ and if R is the gas constant, the work done by the gas is

1. zero
2. $RT_0(V_1/V_0)^\gamma$
3. $RT_0(V_1/V_0-1)$
4. $c_vT_0\left(1-(V_0/V_1)^{\gamma-1}\right)$
5. $RT_0\ln(V_1/V_0)$

Thermodynamics$\Rightarrow$}Work

The work done by a gas in an isothermal expansion is related to the log of the volumes. If one forgets this, one can quickly derive it from recalling the definition of work $W=\int P dV$ and the ideal gas law equation of state $PV=nRT \Rightarrow P=nRT/V$.

One has $W=\int nRTdV/V = nRT ln(V_1/V_0)$. For 1 mole, one has $n=1$, which yields choice (E).

(And the condition for isothermality $P_1V_1=P_0V_0=nRT_1=nRT_0$ allows one to change the argument in the log.)

Alternate Solutions
 djh1012014-09-02 13:56:24 Solution by elimination: -Eliminate A immediately because change in volume requires work to be done. -Work should be positive when volume increases, negative when volume decreases, and 0 when there is no change. Eliminate B. -The process is reversible, so switching V1 and V0 should give an answer with the same magnitude but opposite sign. This only leaves E. You could actually eliminate all but E with this step alone.Reply to this comment
djh101
2014-09-02 13:56:24
Solution by elimination:
-Eliminate A immediately because change in volume requires work to be done.
-Work should be positive when volume increases, negative when volume decreases, and 0 when there is no change. Eliminate B.
-The process is reversible, so switching V1 and V0 should give an answer with the same magnitude but opposite sign. This only leaves E. You could actually eliminate all but E with this step alone.
greatspirits
2012-11-01 15:56:50
I was actually way too zombified from studying by the time I got to this problem, I couldn't even think, usually this would be easy for me, but since I had done the work before, I just remembered that there was a $ln$ in the expression, so E.
bkardon
2007-10-05 11:00:46
Wait - so the information about $\frac{c_p}{c_v} = \gamma$ was a total red herring?
 ewhite22007-10-27 12:50:31 yup
 FortranMan2008-10-01 19:23:02 they're trying to confuse you into thinking that the adiabatic exponent or the compression ratio is involved, when both really aren't since neither gamma is defined with the specific heats. And considering gamma is normally used as an exponent, why would you divide the specific heats anyway?
 Nathan_Grey2010-01-31 22:30:13 FortranMan: I understand that gamma is not needed for this problem, but gamma is customarily the variable used to define the ratio of specific heats. It is also related to the degrees of freedom by: $\frac{c_p}{c_v}$=gamma=$\left(1+2/f\right)$
 kaic2013-10-11 15:46:49 to clarify, that's $\gamma$ = $\frac{f + 2}{f}$
jax
2005-12-05 19:41:50
On my exam the ln (natural log) looks more like $\ell n$ . Is that just a misprint? At first I crossed out this answer choice because I didn't know what the variable $\ell$ was and I thought ETS was trying to trick people into thinking it was a ln. None of the other answer choices worked though so I had to check here. Does anybody else notice this on their exam?
 yosun2005-12-06 19:21:33 jax: it's not a misprint. i have the curly-$\mathcal{l}$ for ln on my GR9277 exam too. you shouldn't cross out choices based on funny looking print. moreover, you should understand the difference between isothermal/baric/sentric/choric work for an ideal gas.
 jax2005-12-06 21:06:06 I only have a curly l in one ln on the exam, all the other lns look normal, so I guess it IS a misprint (the fault of ETS). I actually went back when I was doing the problem and I noticed that all of the other lns in other problems had a regular L. weird. I do understand the problem though. Just confused about the $\ell$.

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