A thin partition divides a container of volume V into two parts. ...

Question

A thin partition divides a container of volume V into two parts. One side contains nA moles of gas A in a fraction fA of the container; that is, VA = fAV. The other side contains nB moles of a different gas B at the same temperature in a fraction fB of the container. The partition is removed, allowing the gases to mix. Find an expression for the change of entropy. This is called the ,entropy of mixing.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, determine the equation for the entropy change. In the given scenario in a greenhouse with a total volume V. Two different gasses P and Q are initially separated by a transparent curtain gasses mix and disperse when the curtain is removed at the same temperature gas P produced by plants occupies a volume fraction of Q subscript P and contains N subscript P moles. While gas Q form by the decomposing organic matter occupies a volume fraction Q subscript capital Q and contains N subscript Q moles. OK. So our end goal is to determine the equation for the entropy change in our given scenario. So the greenhouse condition. OK. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A is delta S is equal to NP multiplied by R multiplied by Ln of V, divided by QP multiplied by V plus NQ multiplied by R multiplied by the natural log of V divided by Q subscript capital, Q multiplied by VB is delta S is equal to V divided by QP multiplied by B plus NQ multiplied by Rlnov, divided by Q subscript capital Q multiplied by V C is delta S equals NP multiplied by R multiplied by L and uh V divided by QP multiplied by V minus NQ multiplied by R multiplied by L NAV, divided by Q subscript capital QV and D is delta S is equal to NP multiplied by R multiplied by the Ln of V divided by QP multiplied by V minus NQ multiplied by R multiplied by LNV. OK. So first off, let us model the gasses P and Q as ideal gasses and that the temperature will remain the same. So the temperature remains constant if it remains the same. So let us also assume that the only thing that is affected by the removal of the curtain is that the volume V that the gasses fill the, fill up the greenhouse or fill up inside the greenhouse. So when the curtain is removed, each gas gets larger in volume than it was before due to the number of micro states. Thus the entropy will increase. So the total change in entropy can be written as and let's call it equation one. So the total change in entropy is delta S, it's equal to delta sp plus delta SQ. OK. Where delta SP is the change in entropy of the gas P and delta sq equals the change in entropy of gas Q. So the change of entropy for an ideal gas is delta S is equal to the number of moles multiplied by CV, multiplied by Ellen of the final temperature divided by the initial temperature plus the number of moles multiplied by R multiplied by Ellen of the final volume divided by the initial volume. OK. Or CV is the heat capacity and capital R is the universal gas constant. OK. So since the temperature remains the same Ln of the final temperature divided by the initial temperature is equal to Ln of one is equal to zero. Therefore, the change in entropy of a gas due to the change in its volume is delta S is equal to N multiplied by R multiplied by Allen of the final volume divided by the initial volume. OK. So for gas, P delta SP is equal to NP multiplied by R multiplied by Allen of VFP. So the final volume of gas P divided by the initial volume of pressure P. And then for gas, Q delta S Q is equal to NQ multiplied by R multiplied by Ln of VF the final volume of gas Q divided by the initial volume of gas Q. OK. So we can rewrite equation one using the equations that we just wrote for gasses P and Q. And let's call this equation two. So equation two, we can write it as delta S is equal to NP multiplied by R multiplied by Ln of the final volume of P of the pressure. So the final volume of the gas P divided by the initial volume of gas, P plus NQ multiplied by R multiplied by allen of the final volume of gas Q divided by the final volume. Sorry the final volume of gas Q divided by the initial volume of gas Q OK. So the volume fraction of gas P is QP. So the initial volume P equals the initial volume of P is equal to QP multiplied by V. And the volume fraction of gas Q is Q subscript capital Q. So the initial volume of Q is, this Q is equal to the initial volume of the gas, Q is equal to Q subscript capital Q multiplied by V and the final volume of the gas for P and Q that can occupy the greenhouse after the removal of the curtain is the, so the final volume of the gas P is equal to the final volume of gas, Q is equal to V. So we can rewrite equation two to state delta S is equal to NP multiplied by R multiplied by Ln of V divided by QP multiplied by V plus NQ multiplied by R multiplied by Ln of V divided by Q subscript capital QV. And this is our final answer. We did it. OK. So let's look at our multiple choice answers to see what the correct answer. Is the correct answer. Turns out to be the letter A delta S equals NP multiplied by R multiplied by L NAV, divided by QP multiplied by V plus NQ multiplied by R multiplied by L NAV divided by Q subscript capital QQ multiplied by V. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Um.

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