## Quiz 1 Preparation: Question #3 - Conflicting Information? [ENDORSED]

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Michael_Johanis_2K
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### Quiz 1 Preparation: Question #3 - Conflicting Information?  [ENDORSED]

#3. "If 2.00 mol of an ideal gas at 300. K and 3.00 atm expands isothermally and reversibly from 6.00 L to 18.00 L and has a final pressure of 1.20 atm, what is q,w, and DeltaU?"

Can anyone clarify what I am doing wrong?

w = -nRT ln (V2/V1) OR
w = -nRT ln (P1/P2)

ln(V2/V1) and ln(P1/P2) have to be equal to each other, since we should be able to solve the question either using the two pressures or the two volumes.

ln (18.00 L / 6.00 L) = 1.099
ln (3.00 atm / 1.20 atm) = .916

I think that we can obtain different values of w (and therefore heat, since q = -w in isothermal reverse expansion) depending on whether we use pressure or volume for the ln( ) part. What is the problem here?

The question states that this is isothermal, reversible expansion, so it is a downward curve on a P-V graph, and PV is constant. I don't think that the graph could consist of a vertical line and a horizontal line (This would be irreversible expansion, right?).

Thank you!

Chem_Mod
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### Re: Quiz 1 Preparation: Question #3 - Conflicting Information?

Typically the ratio of pressure and volumes are equal, but as we can see in this problem they are not. For reversible, isothermal problems use the first equation (with volume). Here is the work for the solution.

Chem_Mod
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### Re: Quiz 1 Preparation: Question #3 - Conflicting Information?

Final pressure in the question should be 1.00 atm. The above solution is correct as is.

A bit early, ... this will be covered in about 2 weeks time in class.

sboutros2B
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### Re: Quiz 1 Preparation: Question #3 - Conflicting Information?

good evening! so just to clarify how would we know when to use whether pressure or volume in the ln part of this equation?

samuelkharpatin2b
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### Re: Quiz 1 Preparation: Question #3 - Conflicting Information?

Sboutros2b, we use the equation for work which includes final and initial volume when the pressure changes. If the pressure is constant, then we simply use w=-Pdelta(V) to calculate work. And since the question states that the process is isothermal, we know that the internal pressure is zero. (For all isothermal processes this is the case). And since we know that fact, q turns out to be the negative/opposite of the work.