A reaction is carried out in a cylinder fitted with a movable pis...

Question

A reaction is carried out in a cylinder fitted with a movable piston. The starting volume is V = 5.00 L, and the appa- ratus is held at constant temperature and pressure. Assum- ing that ∆H = -35.0 kJ and ∆E = -34.8 kJ, redraw the piston to show its position after reaction. Does V increase, decrease, or remain the same?

Accepted Answer

hi everyone for this problem it reads consider a gas sample is contained in a cylinder with a movable piston shown below a process with entropy change equal to negative 24.5 kg jewels and internal energy equal to 25 kg jewels is done on the gas sample at constant temperature and pressure. Which image corresponds to the state of the gas after the process. So the question that we want to answer is which are the images corresponds to the state of the gas after the process. So we have a gas sample and a movable piston were given the entropy change and the internal energy change along with pressure. So these are the three pieces of information were given pressure entropy change and internal energy change. So the equation that we're going to want to use to solve this problem is entropy change is equal to internal energy change plus pressure times change and volume. And for the question we're being asked about the image after the process. So that means we want to know what's the change in volume of the gas after the process. So the variable that we want to isolate here is change and volume. Okay, so let's go ahead and make this on one side of the equation. So we're gonna have pressure times volume is equal to we're gonna move our internal energy change to the other side. So when we do that we're going to get entropy change excuse me we're going to get yes entropy change minus internal energy change. Okay so we want to isolate change in volume. So let's divide both sides by pressure. So once we isolate our change in volume we get change in volume is equal to entropy change minus internal energy change over pressure. And these three values were given in the problem. So let's just write them out pressure we're told is 1.7 atmospheres. Entropy change is equal to negative 24.5 kg jewels. And the internal energy change is equal to negative 25 kg jewels. So we have everything we need. So let's just go ahead and plug these values in. So we get our change in volume is equal to our entropy change is negative 24. kg jewels and this is going to be minus internal energy change which is negative 25 Killer jewels and this is all over pressure and our pressure is 1.7 atmospheres. So let's go ahead and do this calculation. And when we do this calculation the answer that we're going to get is 0. kg joules per atmospheres. Okay so our unit is in kilo joules per atmosphere but we want this and volume we want to convert this so we want to go from killer jewels per atmosphere to leaders And one of the conversions that we're going to need to know in order to do this is that one atmosphere leader is equal to 101.3 jewels. Okay so here we see we're in kila jewels. So let's start off by going from killer jewels, two jewels. Alright, so in one kill a jewel there is 1000 jewels. So our units for killer jewels cancel. And now we're in jewels per atmosphere. But we want to go to Leaders. All right. So what we're going to do then is use this conversion right here. All right. And so we're going to write in one atmosphere leader there is 101.3 jewels. Alright, so looking at our units, atmospheres cancel Jules cancel. And we're left with the unit of leaders which is what we want. So let's go ahead and do this calculation we're going to take .2941 and multiply it by and then divide it by 101.3. Okay, so once we do this calculation, the answer that we're going to get is 2.9 liters. So we can say here that are change in volume is equal to 2.9 liters. And what this means is the volume increases. So looking at the image after the process, the volume increases. So looking at our images, Image one compared to our original image, we have the volume increasing. Image B It looks like the volume is the same and image. See it looks like the volume decreased. So our correct answer then for this problem is going to be a Alright, and this corresponds to the volume increasing, which is what we identified. Okay, So that is it for this problem, The image that corresponds to the state of the gas after the process is image A. And this is our final answer. That is it for this problem. I hope this was helpful.

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