(II) Two samples of an ideal gas are initially at the same tem...

Question

(II) Two samples of an ideal gas are initially at the same temperature and pressure. They are each compressed reversibly from a volume V to volume V/2, one isothermally, the other adiabatically.

(b) Determine the change in entropy of the gas for each process by integration.

Accepted Answer

Welcome back. Everyone. In this problem, a physicist takes two pistons that initially contain the same amount of hydrogen gas at the same temperature and pressure to examine the behavior of the hydrogen. He compresses them from a volume of V to a volume of 1/5 of V reversibly. He compresses one of the pistons isotherm and the other is compressed adiabatic using integration, calculate the change in entropy for each process. A says that change in entropy for the adiabatic process is negative and R multiplied by the natural log of 1/5. While for the isothermal process, it's negative and are multiplied by the natural log of five B says they are negative N RT Ln 1/5 and NRLN 25 respectively. C says they are zero and negative NRLN five respectively. And D says they are zero and NTLN five respectively. Now let's start by looking at the change for the adiabatic process. What do we know there? Well, generally we know OK, we know that our changing entropy delta S is equal to the integral of DQ divided by T OK. Or the reciprocal of T one divided by T with respect to Q where Q is the heat received by the system and T is the temperature in Kelvins. Now, for the adiabatic process, we know that no heat is exchanged in an adiabatic process. That means that, so let me make a note here so far in our first scenario, we know for an adiabatic process, then the change in heat equals zero. So that tells us then that the change in entropy therefore is going to be equal to zero. No, let's next move on to the isothermal process. Let's call that part two. OK. Now what do we know? Well, the word isothermal itself tells us that it's going to be at a constant temperature and for the isothermal process. So for the isothermal process, OK, then the change in internal energy is going to be the difference between the heat and the heat received by the system. Sorry. And the work done there is no change in internal energy. So that means zero equals Q minus W which tells us then that the work done is equal to the heat received by the system Q. Now applying what we know about our change in entropy, this tells us then that our change in entropy for the isothermal process. OK is going to be equal to, as we know the integral of DQ divided by T or DQ over T which is one divided by T multiplied by the integral of DQ. Because remember we said the temperature is constant for our isothermal isothermal process. So T is constant. So in that case, then if we integrate DQ that tells us that our change in entropy for the isothermal process is going to be equal to delta Q divided by T. Now, what do we know about delta Q? Well, we also know that delta Q change in heat is going to be equal to NRT multiplied by LNVF divided by V. OK, where VF is the final volume and V is the initial volume. Remember earlier, we were told that our initial volume was V OK? And our final volume was 1/5 of VEVF is 1/5 of V. So now that tells us that if we put that into our formula, OK, then we're really saying that the change in entropy for isothermal process is going to be equal to NRT divided by T multiplied by the natural log of our final volume. 1/5 of V divided by our initial volume which is V or we can write that as V fifth multiplied by one divided by V. Now we know here that RV is cancer or A T is cancer. And now we're left with that change in entropy to be equal to N or NRLN of 1/5. Now we can rewrite the natural log of 1/5 using the rule that tells us, let me put it over here that the natural log of a reciprocal is equal to the negative value of the natural log of that number. So really, then we can rewrite the natural log of 1/5. OK. Let me write that properly here as the natural log of negative Ln five. So that is that change in entropy four or isothermal process then can eventually be written as negative Nr multiplied by L five. Thus, that tells us the change in entropy for our isothermal process, which as we said is negative NRLN five. And the change in entropy for our adiabatic process is zero. Therefore, C is the correct answer. Thanks for watching everyone. I hope this video helped.

(II) Two samples of an ideal gas are initially at the same temperature and pressure. They are each compressed reversibly from a volume V to volume V/2, one isothermally, the other adiabatically.
(b) Determine the change in entropy of the gas for each process by integration.

Key Concepts

Entropy

Isothermal Process

Adiabatic Process

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