Entropy Calc: Solving for 100 Mol Gas

[ljeonjko]

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 :)

 

[Chestermiller]

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?

 

[ljeonjko]

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.

 

[Chestermiller]

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.

 

[Chestermiller]

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?

 

[ljeonjko]

Starting from

dQ=dE+dW

it's easy to obtain

S=nC_vln(T_f/T_i)+nRln(V_f/V_i)

using the three other equations.

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

The main question still remains whether or not the V_f1 and V_f2 are the same as V_i1 and V_i2 respectively. If they are, the result is 0.777 J/K and if they are not, the result is 1153.77 J/K.

 

[Chestermiller]

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