# Homework Help: Thermodynamic equilibrium of ideal gas and mechanical system

1. Sep 7, 2014

### Math Circles

1. The problem statement, all variables and given/known data

Hi I am new to Physics Forums and desperately need some help! This is my first class back since undergrad and my math skills have gotten a little rusty. I think that's where I am failing. Okay here goes:

http://tinypic.com/r/2zqf04k/8

There is a 4 subsystem Isoentropic system filled with ideal gases. It contains: a heat reservoir (r), which contributes heat to systems 1 and 2. Systems 1 and 2 which have a movable wall between them but do not share temperature or concentration. And the system (m) which is a pulley system with a weight on the end with mass "m" and is a measurable height off the ground.

let "s" be the cross section of the wall separating systems 1 and 2.

(1) Derive the conditions for equilibrium of only systems r, 1 and 2.

(1A) Now Derive a relationship between N1,N2 and the equilibrium volumes V1°,V2°.

(2) Derive the conditions for equilibrium of all 4 systems.

(2A) Using the initial conditions found in 1A derive a relationship between m and the new equilibrium volume of system 1.

(3) Linearize the relation between m and V derived in 2A by assuming that V1=V1° +ΔV1. Express as m=αΔV1

2. Relevant equations
F=MA
F=P(area)
PV=NRT
E=cNRT
E(S,V,N,T)

3. The attempt at a solution

(1) Since the system is isoentropic, using Esys=Er+E1+E2 I took the partial derivatives of E with respect to S,V,N and T for each system and then simplified to:

0=((dE1/dS1/)v1N1 - (dEr/dSr))

0=((dE2/dS2/)v2N2 - (dEr/dSr))

0=((dE1/dV1)N1S1 -(dE2/dV2)N2S2)

Then from the diagram:

T1(S1,V1,N1)=Tr=T2(S2,V2,N2)
P1(S1,V1,N1)=P2(S2,V2,N2)

(1A)

PV=NRT substituted into P1(S1,V1,N1)=P2(S2,V2,N2) gives: (N1R1T1/V1)=(N2R2T2/V2)

The T1 and T2 can be substituted as Tr since T1(S1,V1,N1)=Tr=T2(S2,V2,N2)

So I get: N1/V1=N2/V2

Then since the total volume of system 1 and 2 is constant:
V°(1,2)=V°1+V°2

(2)

Esys= Er +Em + E1 +E2 I took the partaials of Er,E1 and E2 with respect to S,V,N like before and then Em with respect to Height and S. Is that correct?

I got back:

0=((dE1/dS1/)v1N1 - (dEr/dSr))

0=((dE2/dS2/)v2N2 - (dEr/dSr))

0=((dE1/dV1)N1S1 -(dE2/dV2)N2S2)

0=((dEm/dSm)mH-(dEr/DS4))

0=(dEm/dH)m,Sm

P°(1,2)= P1 +P2

V(1,2)= V1+V2

(2A)

This is where I really get stuck. I am very tempted to set up the equation F(1,2,m)= F1 -F2 +Fm based off of free body diagrams I drew in the system. From there using F=MA for the mass system and F=P(area of the wall denoted by s.)

Substituting in I get: 2(P1)(s)- (P°(1,2))(s)+mg

Substituting in PV=NRT and V1= V°+ΔV and T1=T2=Tr I get : TrR((2N1/V°)+2(N1/ΔV)-(N°1,2/V°(1,2))+mg

Am I on the right track here? Should I have used and Energy equation instead? I tried that with combining PVT=NRT with E=cNRT and E=mgH and got an Equation:

E= cP1V1 +cP2V2+ mgH
and after taking the derivative an answer of m=-2ΔP1ΔV

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Sep 8, 2014

### Staff: Mentor

Hi Math Circles. Welcome to Physics Forums.

I've looked over your problem, and feel that you are misinterpreting it and making it much more complicated than it actually is. My understanding of the problem statement is that, in the initial equilibrium state of the system, the mass m is not attached, and that, during the deformation, the mass m is gradually lowered to its final equilibrium position. The fact that it is lowered gradually follows from the condition that the overall system is isentropic. This implies that the deformation is carried out reversibly, since, if the mass were just released without support, the deformation in the enclosure would be irreversible (and the change would not be isentropic).

Since the boundary between the reservoir and the two gas chambers is diathermal, what does that tell you about the temperature of gases in the two chambers compared to the temperature of the reservoir? Without the mass m attached in part 1 of the problem, how do the pressures of the gases in the two chambers compare? The answers to these two questions should give you enough information to address part 1A.

In part 2, the gases in the two chambers undergo volume changes. What are the final temperatures of the gases in the chambers (considering that they each remain in contact with the reservoir)? If V1 is the final volume of chamber 1, what is the final volume of chamber 2? From these results and the ideal gas law, what are the final pressures in the two chambers? If you do a force balance on the piston between the two chambers, what does this force balance give you for the difference between the two pressures? In terms of V1, what are the pressures in the two chambers individually?

Chet

3. Sep 8, 2014

### Math Circles

I feel pretty comfortable with part 1, but for part 2:

If V1 is the final volume of chamber 1 than the final volume for chamber 2 is V(1,2)°-P1 or V1°+ΔV1.

or P2=N1RT1/V1+ΔN1RΔT1/ΔV1

Force equation Ftotal= -F1+F2-Fm

substituting F=PA and F=ma

(-n1RT/V1)A+(n2RT/V1°+ΔV1)A-mg

or RTA((-n1/V1)+(n2/1°+ΔV1))-mg

But i don't understand how that is linearized so that it can be expressed as m=αΔV. I know because ΔV is small it can be considered the same of the 1° taylor expansion (essentially indistinguishable from its tangent line,) but my answers never come out clean.

4. Sep 8, 2014

### Staff: Mentor

I think you have the right idea, but I'm having lots of trouble reading your equations and notation. Please read the tutorial on Latex equation editor in Physics Forums.

At final equilibrium, the net force is zero, so the acceleration is zero. Also, for the volumes:
$$V_1=V_1^o+ΔV$$
$$V_2=V_2^o-ΔV$$
You also need to make use of the initial condition when mg = 0, and subtract that force balance equation from the final force balance equation. This will lead to an equation that you can linearize, and solve for ΔV, with ΔV being proportional to m.