(a) What sign for Δ𝑆 do you expect when the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm?

(b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process.

(c) Which of the following statements about this process are true? (i) The entropy change you calculated will be the same for at any other constant temperature. (ii) The value of Δ𝑆 you calculated is valid only if the compression is done irreversibly. (iii) If the number of moles of gas being compressed were decreased by a factor of three, the entropy change would increase by a factor of three.

Video transcript

Hey everyone, we're told that the pressure of 0.750 mole ideal gas at 330K. With an initial pressure of 0.800 atmospheric pressure is increased so thermally a predict the sign of the change in entropy for this process be what is the entropy of change if the pressure on the gas is 1.50 atmospheric pressure. As we've learned, we know that Boyle's law states that P one times V one equals P two times V two, which essentially tells us that as our pressure increases, our volume decreases. And in our questions them, we can see that we had an initial pressure of 0.800 atmospheric pressure and this was increased to 1. atmospheric pressure. So because our pressure increased, this means that our volume decreases and we have ice a thermal compression. So the smaller volume means smaller space for the particles to move. So answering a this tells us that our change in entropy is going to be negative. Now looking at B were asked what is the entropy of change? If the pressure on the gas is 1.50 atmospheric pressure, we can go ahead and answer this question by using the following formula. So the entropy of change for our system is equal to N times are gas constant. R times the natural log of our volume two divided by our volume one. Now in order to get volume two divided by volume one, we know that P one times V one equals P two times V two. Now in order to isolate volume two divided by volume one. We end up with pressure one divided by pressure to so solving for that ratio, we can go ahead and plug in our values. Our pressure one was said to be 0.800 atmospheric pressure. And we're going to divide that by our pressure to which was 1.50 atmospheric pressure which gets us to A value of 0.5333. And this is going to be the value for Volume two divided by Volume one. Now let's go ahead and use that value to solve for the change in entropy of our system. So we got 0.750 mole times r gas constant which is 8.314, joules divided by mole times kelvin Times Our Natural Log of 0.5333. And this gets us to a change in entropy of negative 3. jewels. Over Kelvin. Now looking at C it is not necessary to specify the temperature since the process his eye. So thermal. So essentially we have a constant temperature and these are going to be our final answers. Now, I hope that made sense. And let us know if you have any questions