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Topic: calculating work, dE, and dS  (Read 5899 times)

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Offline a confused chiral girl

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calculating work, dE, and dS
« on: April 13, 2007, 03:24:06 AM »

I have a question that I am pondering on: One mole of an ideal monatomic gas (Cv=3R/2) that is intially confined to a vewssel of volume 0.5L expands reversibly at constant pressure such that the final volume is 1L. For this expansion, I caluclated that work is less than 0 by ---> w = -nRTln(Vf/vi) = -nRTln(2) = a "-" number.

so to calculate delta E, I used this equation dE = q + w , and from above w is a negative number, thus dE = q + (-w). but the answer is dE > 0, I am stuck on how this is so. then to calculate the dS, I used dS = nRln(2) , and in the question it says 1 I assumed n = 1. thus I got dS = Rln(2). but the correct answer is dS = 2.5Rln2. I am confused about this.

can anyone please explain? thanks!

Offline oceane

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Re: calculating work, dE, and dS
« Reply #1 on: June 17, 2007, 05:03:51 PM »

Well I got dS=2Rln2 which seems to be still not the right answer but mine differ from yours because you haven't converted the units of litre into the standard units of m^3.

i.e 0.5L is (0.05 m)^3    and 1L is (0.1 m)^3

Offline Hunt

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Re: calculating work, dE, and dS
« Reply #2 on: June 20, 2007, 05:07:57 PM »

dS = dQ / T = nCp dT/T ( const Pressure ) ==> delta S = nCp Ln T2/T1 = nCp Ln (V2 / V1) = 5/2 R Ln2 = 2.5 R Ln2 because Cp + Cv = nR

As for dE , ofcourse it is positive. dE = nCvdT for any ideal gas. As V increases so does T. Hence delta T > 0. so delta E > 0.

Offline renai

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Re: calculating work, dE, and dS
« Reply #3 on: July 24, 2007, 11:48:18 PM »

Remember, your expansion is isobaric.  Now you should have had a -ve work answer because the system is losing energy by doing expansion work it's just that the heat it is receiving is positive and greater in magnitude than the work it does.  One thing to note is that your work calculation is wrong because on integrating you're removing T from the integral which is incorrect.  T increases in this case so you can't get -nRTlnV2/V1.  If you really want to know the work term, since P is constant, simply plug the initial or final (T,V) values into the equation of state P=nRT/V which will effectively give you the external pressure (you said this process was reversible) and multiply by the delta V (but put a negative sign in front of that).  Now you need the heat term, Q.  which under isobaric conditions is equal to delta H. which equals Cp(T2-T1).  This value will be positive and should be greater in magnitude than the -ve work term you get.  Add the 2 together and voila, you'll get your positive delta E answer which make sense since the internal energy of an ideal gas is solely a function of temp. and the temp. of your gas must go up since P = nRT/V, V is going up and P is not changing.


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