CALC A lonely party balloon with a volume of 2.40 L and containin...

Question

CALC A lonely party balloon with a volume of 2.40 L and containing 0.100 mol of air is left behind to drift in the temporarily uninhabited and depressurized International Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 m^3. Calculate the entropy change of the air during the expansion.

Accepted Answer

Hey everyone in this problem, we have a container volume 50 centimeters cubed. Okay, It's connected with a short tube fitted to a valve with a centimeter cube container attached that's evacuated using a vacuum pump. The 50 centimeter cube container has one mole of oxygen at 290 Calvin. When the valve gets open, the oxygen is going to undergo free and ice of thermal expansion and fill both containers. And we are asked to find the entropy change DELTA S. During this free expansion. So let's recall the entropy change. Delta S. Is equal to Q. Divided by T. So cue the quantity of heat T. The temperature. Now We don't know Q. We do know T. Okay, we're given the temperature 290 Calvin. But what do we know about this quantity Q. We're told this is an icy thermal expansion. We have oxygen. So we can consider this as an ideal gas. Okay, so I use the thermal expansion with ideal gas. This tells us that delta U. The change in internal energy is going to be zero. If the change in internal energy is zero, we get that Q. Is equal to the work. W. So in order to find delta S. The change in entropy that we're looking for, we need to find Q. First. And what we know from this being ice thermal than ideal gasses that Q. Is equal to the work. W. So instead of finding Q. We can find the work W. Which might be easier in this case. Okay, Alright, so let's start with work. Okay, let's recall that we can write the work when we're talking about two different volumes is equal to n the number of moles are the gas constant. T the temperature times Lawn V. two over v. one. Alright, well, this is good. We know end the number of moles were given in the problem are as a gas constant. We know that we know the temperature and we're told the two volumes. Okay, So we can go ahead and find the work straight away and we're given Is one more. Ok. The number of moles is one. So we get one more R is the gas constant which is 8.314. Okay. With the units jewel per more Calvin. We have the temperature T 290 Calvin. Many times lawn Of the ratio of the volumes. Okay, now let's think about this. The first volume we know is 50cm cubed. Okay, that's where we're keeping the initial oxygen. When we open that valve to the container of centimeters cubed, the oxygen is going to fill both containers. Okay, so the final volume or V two is going to be the sum of both of those volumes. Okay, so it's going to be the 50 centimeter cube container plus the 100 centimeter cube container. Okay. And the initial volume V one is just again that 50 centimeter cubed container. All right. So, if we work this out on our calculators, we are going to get to work 2648.82 jewels. Now, remember because we are so thermal ideal gas. We said Q. Is equal to W. So we found this work W This is actually just equal to Q. Okay, so now we have a value of Q. That we can use in our delta S equation to find the change in entropy. Alright, perfect. Let's give ourselves a little bit more room and we're gonna go back to that entropy equation. So delta S quantity we're looking for is equal to Q. 2648.82 jewels divided by the temperature 290 Calvin. And this is going to be equal to 9.13 units jewels for Calvin. There we have it. So the change in entropy is going to be D 9. joules per Calvin. Thanks everyone for watching. I hope this video helped see you in the next one.

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