Calculate the work for each of the following processes beginning with a gas sample in a piston assembly with T=305 K, P=1.79 atm, and V=4.29 L; (a) irreversible expansion against a constant external pressure of 1.00 atm to a final volume of 6.52 L; (b) isothermal, reversible expansion to a final volume of 6.52 L.
What is the easiest way to approach this problem? Should we use the equation w= -nRT*ln(Vf/Vi) or another equation?
Topic 4b-interal energy question #4b.13
isochoric/isometric:
isothermal:
isobaric:
isothermal:
isobaric:
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Re: Topic 4b-interal energy question #4b.13
Postby Chem_Mod » Sat Feb 04, 2023 7:30 am
Hi,
The easiest way to approach this question is to start from the expression dw = -pdV and then integrate over both sides of the equation. In a), you can pull the pressure out, since it is a constant. In b), you'll need to derive an expression for pressure through rearrangement of the ideal gas law. You can plug that into p and then you'll have something as a function of v instead the integral. You can then perform that integration to get the work expression you have listed in your problem. So yes, that expression is correct for this problem. You'll note that it makes sense to use since there is no delta T, and the expansion is performed isothermally (just a good sanity check).
Hope this helps!
The easiest way to approach this question is to start from the expression dw = -pdV and then integrate over both sides of the equation. In a), you can pull the pressure out, since it is a constant. In b), you'll need to derive an expression for pressure through rearrangement of the ideal gas law. You can plug that into p and then you'll have something as a function of v instead the integral. You can then perform that integration to get the work expression you have listed in your problem. So yes, that expression is correct for this problem. You'll note that it makes sense to use since there is no delta T, and the expansion is performed isothermally (just a good sanity check).
Hope this helps!
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