[Stephen Clarke]

Given a sample of nitrogen gas (assume ideal gas conditions), the following conditions were observed inside the container. n = 0.75 moles at 253 K, and pressure = 0.5 atm. Then, an ISOTHERMAL IRREVERSIBLE COMPRESSION on the system forced by a constant Pexternal = 10 atm reduced the initial volume by a factor of 2. Assume diathermal walls on this container. Calculate the entropy change of the system, surroundings and universe.

ATTEMPT:

The entropy change of the system is under ISOTHERMAL conditions, so, dT = 0, and,

dS = nRln(P2/P1)
dS = (0.75 mol)(8.314 J/Kmol)ln(1/2)

However, I am having a difficult time conceptualizing the entropy change of the surroundings. This is obviously an integral part of computing the entropy change of this particular universe.

I'll appreciate any feedback! Thank you.

[mjc123]

Try starting with ΔU = q + w.

(To be pedantic, ΔS = nRln(P₂/P₁). d denotes an infinitesimal change; Δ a finite difference.)

[orthoformate]

Hi everyone,

I'm a student also trying to figure out this problem.

using dU=TdS-PdV we derived ΔS=nRln(V₂/V₁)
giving -4.14 J/K for the system... is this correct?

We also tried:

ΔS = nCp*ln(T2/T1) + nR*ln(V2/V1) assuming PV/T = nR = constant therefore P1*V1/T1 = P2*V2/T2
plugging that in to ΔS eqn above you get n*Cp*ln(P2/P1) + n*Cp*ln(V2/V1) + nR*ln(V2/v1)
we know for diatomic ideal gas molecules that Cp = (7/2)*R, then plug in the values for V2, V1, P2, P1, R, and n
and you should get ΔS to equal 45.93 J/K

We are unsure how to calculate the entropy of the surroundings.

[mjc123]

using dU=TdS-PdV we derived ΔS=nRln(V2/V1)
giving -4.14 J/K for the system... is this correct?

Approach is correct, but I disagree with your number. How did you get it?

ΔS = nCp*ln(T2/T1) + nR*ln(V2/V1)

It's isothermal. What is T2/T1?

n*Cp*ln(P2/P1) + n*Cp*ln(V2/V1) + nR*ln(V2/v1)

What is the relation of P2/P1 to V2/V1?

and you should get ΔS to equal 45.93 J/K

How??? (note: P2 is not 10 atm!!!)

[orthoformate]

Hi Mjc123,
Thank you very much for the response! We are all putting our heads together here trying to solve this question


ΔU=TΔS-PΔV rearranges to TΔS=PΔV (as ΔU=0 in isothermal operations)

ΔS=PΔV/T so dS=PdV/T

rearranging to remove pressure

dS=(nRT/VT)dV

Temperature cancels, and the formula is integrated

ΔS=nRln(V₂/V₁)

we know that the initial volume has been reduced by a factor of 2 so V₂=1 and V₁=2

plugging in:

ΔS=nRln(1/2)

ΔS=0.75*8.314*-0.69

ΔS=-4.3 J/K or 1 eu (entropy unit)

where have I gone wrong?

[EngRPI]

Something to point out...the change of entropy for the system and its surrounding is always POSITIVE in an irreversible process.

[mjc123]

Something to point out...the change of entropy for the system and its surrounding is always POSITIVE in an irreversible process.

Change of entropy of the universe (system plus surroundings) is positive. Not necessarily each individually.

where have I gone wrong?

-4.3 J/K is right. Where did you get -4.14? (And 1 e.u. = 4.184 J/K/mol - IUPAC Gold Book)

[orthoformate]

Thank you! we must have just have made a small error somewhere the first time.

Now we are considering the ΔS of the surrounding. We know Q.

Q=W
Q=PdV
Q=nRT/VdV
Q=(0.75*8.314*253)ln(1/2)
Q=-1093 J

This heat is created by the compression, but is transferred out of the system in order to remain isothermal.

dS=dQ/T

so we need to know the temperature of the surroundings, which should be constant.

[mjc123]

What is the work done on compression by a constant external pressure?

[orthoformate]

What is the work done on compression by a constant external pressure?

Since it is a compression, the pressure that we should be considering is the internal pressure of the system, which is not constant. The external constant pressure does work on the system with an initial internal pressure of 0.5atm (55,000 Pa).

Since P₁V₁=P₂V₂ at constant temperature, then the ending pressure of the system must be 1.0atm (110,000 Pa).

We considered that since the compression is irreversible, then perhaps the compression is instantaneous, giving:

W=p(int)dV
W=55,000Pa*0.0139 m^3
W=764 J

either of these values divided by the temperature of the surroundings, should give the surroundings entropy change

[mjc123]

Not so. The irreversible work done by a constant external pressure is P_extdV. (See here for a discussion: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/ ) If you think about it, it must be greater than the reversible work, so that the entropy change of the surroundings is greater than that of the system (and of opposite sign) and ΔS_uni > 0.

[orthoformate]

Hi mjc123,

So when you instantaneously apply 10 atm of pressure to the system at 0.5 atm of pressure the viscous damper of the system fluid absorbs energy and dissipates it as heat. Simplifying the calculation you can write the internal final pressure (1.0 atm of 101,325 Pa) multiplied by the change in volume (0.0139 cubic meters) to give the isothermal irreversible work performed by the surroundings on the system.

101,325 Pa × 0.0139 m³= 1408 J

Is this correct?

[mjc123]

No!!! Did you read what I wrote? "The irreversible work done by a constant external pressure is P_extdV." So W = 10 atm*ΔV.
If you move a force P*A through a distance dx, how much work is done? Where does that energy go?
Oh, and I get ΔV = 0.0156 m³ (0.75*8.314*253/101325)

[orthoformate]

Ahh Sorry.

I just have a hard time wrapping my head around the fact that the internal pressure is not a factor in this work calculation.
So as you said P_extdV for this work calculation is W = 10 atm*ΔV

W=1,013,250 Pa*0.0139 m³=14,084.175 J

This work is being done on the system, but since the system is isothermal and U=0 all work done is transferred to the surroundings as heat, so surroundings are receiving 14,084.175 J of heat.

So using ΔS= ∫PdV/T we could calculate the change in entropy of the surroundings?

[mjc123]

In this case, yes. More generally ΔS_surr = -Q_sys/T_surr
But I still disagree with your value of ΔV. How did you get it?