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Entropy calculation

  • Thread starter ljeonjko
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Homework Statement


Two isolated containers, of volumes V1 and V2, enclose ideal single atom gas at the same pressure p. The number of particles in each container is equal, the temperature of gas in container one is T1=293K and the temperature of gas in container two is T2=308K. An equilibrium is acheived by connecting the two containers. Calculate the entropy of the system if each container initially contained 100 mol of gas.

Homework Equations


dQ=dE+dW

dS=dQ/T

dE=3/2 nRdT

pV=nRT

The Attempt at a Solution



Basically, I've solved the problem, with the result of (delta)S=1153.77 J/K. Mathematically speaking, it checks out, but I doubt, along with a few of my colleagues, that the result is much too great. I've also searched the wikipedia for the order of magintudes which shows that the standard entropy of 1 mole of graphite is 5.74 J/K. This means that, for 200 moles of the gas, the result should be more or less accurate.

I'd greatly appreciate if anyone could double check this. Thank you :)
 
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Answers and Replies

  • #2
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In terms of the initial pressure p, what did you get for the initial volumes V1 and V2? What was the final temperature? In terms of p, what was the final pressure?
 
  • #3
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Can you please just check and see if the result matches. The concern is whether or not the volumes "mix", meaning V1 becomes V1+V2 just like V2 becomes V1+V2 or if the volumes stay the same, as the problem is not clear on the issue. If we calucate that the voulmes mix, the result should be 1153.77 J/K, otherwise, the result is 0.777 J/K.
 
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  • #4
19,921
4,096
Can you please just check and see if the result matches. The concern is whether or not the volumes "mix", meaning V1 becomes V1+V2 just like V2 becomes V1+V2 or if the volumes stay the same, as the problem is not clear on the issue. If we calucate that the voulmes mix, the result should be 1153.77 J/K, otherwise, the result is 0.777 J/K.
Please show the details of what you did.
 
  • #5
19,921
4,096
Can you please just check and see if the result matches. The concern is whether or not the volumes "mix", meaning V1 becomes V1+V2 just like V2 becomes V1+V2 or if the volumes stay the same, as the problem is not clear on the issue. If we calucate that the voulmes mix, the result should be 1153.77 J/K, otherwise, the result is 0.777 J/K.
Mixing is irrelevant, because it is the same gas in both containers. Did you remember to use Cp and not Cv?
 
  • #6
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Starting from

dQ=dE+dW

it's easy to obtain

S=nCvln(Tf/Ti)+nRln(Vf/Vi)

using the three other equations.

Tf is obtained through Q1+Q2=0. From there, it's just basic math.

The main question still remains whether or not the Vf1 and Vf2 are the same as Vi1 and Vi2 respectively. If they are, the result is 0.777 J/K and if they are not, the result is 1153.77 J/K.
 
  • #7
19,921
4,096
Take as the basis of zero entropy 273 K and pressure p.

Initial entropy gas relative to basis state = ##100C_p\ln(293/273)+100 C_p \ln(308/273)##
Final entropy of gas relative to basis state = ##200 C_p(300.5/273)##

##\Delta S=100C_p\ln(300.5/293)+100C_p\ln(300.5/308)=1.295## J/K

Chet
 
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