Thermodynamics, two equal volumes of gas at STP mix, entropy change?

[krhisjun]

Homework Statement

Calculate entropy change if two equal volumes of gas (1 cm3) at standard temperature and pressure are separated from each other by a partition which is removed to create twice the volume at standard temperature and pressure

Homework Equations

dS = k.N. dV/V + C/T dT
P1V1 = P2V2
PV=nRT

The Attempt at a Solution

Im making the assumption that the two gases are not ideal, this question leads on to an essay on the gibbs paradox and i believe the correct anser should be zero entropy change.
Im struggling to prove this though as the pressure must stay the same before and after, if the pressure stays the same then due to boyles law the volume must stay the same, but in the question it ofcourse states the volume increases to twice its origninal value.

Im sure i must be missing something here?

 

[vela]

The volume of each gas doubles, so the partial pressure of each gas is halved. The total pressure, however, remains 1 atm.

 

[krhisjun]

but how do i go about proving entropy change is zero

 

[vela]

I don't think it is 0, actually. Why would it be?

 

[krhisjun]

As far as I've researched it is infact zero due to the fact that the molecules are indistinguishable from each other, the partition can just be reinserted and the thermodynamic state is the same so it hasnt changed in entropy.

 

[Andrew Mason]

As far as I've researched it is infact zero due to the fact that the molecules are indistinguishable from each other, the partition can just be reinserted and the thermodynamic state is the same so it hasnt changed in entropy.

The molecules in the gas have a larger volume to occupy when you remove the partition. The situation is equivalent to a free expansion since the gas on each side occupies double the volume without doing work. There is no heat flow (both sides at the same temperature). So there is no change in internal energy: dQ = 0 = dU + PdV = dU + 0 => dU = 0;

But the fact that there is no heat flow does not mean there is no change in entropy. That would be the case only if the path taken was reversible and this mixing process is not reversible.

In order to find the change in entropy between two states, you have to first find the reversible path between those two states and then calculate the integral of dQ/T over that path. For the expansion of the gas from V to 2V with no change in U, the reversible path is a quasi-static isothermal expansion. In this quasi-static expansion, work is done on the surroundings (the external pressure is infinitessimally less than the pressure of the gas at all times). This means that heat must flow: From the first law, since dU = 0 => dQ = PdV > 0

Work out that integral for the gas on each side of the partition and that will give you the total change in entropy.

AM

 

[Andrew Mason]

As far as I've researched it is infact zero due to the fact that the molecules are indistinguishable from each other, the partition can just be reinserted and the thermodynamic state is the same so it hasnt changed in entropy.

The question as stated does not say that the gas molecules on either side of the partition are identical. If they are identical, then you are correct.

AM