0.25 mol of a gas are compressed at a constant pressure of 250 kPa from 6000 cm^3 to 2000 cm^3, then expanded at a constant temperature back to 6000 cm^3. What is the net work done on the gas?

Question

Accepted Answer

Everyone in this problem. A 0.32 mole gas sample experiences isobaric compression at 2.2 atmospheres from V equals 5 L to V equals 1.5 L followed by isothermal expansion to the starting volume of 5 L. We're asked to calculate the sum of the work done on the gas. We're given four answer choices all in jewels. Option A 395 option B negative 402 option C negative 378 and option D 410. All right. So we wanna get started. OK? And we want those work that we're gonna calculate to be the sum of the works from these two processes. OK? So we're gonna have the work equal to the work done by the Isobar compression. We're gonna call that WC. OK. So the work due to the compression lost the work from the expansion W OK. So this is the total work we're looking to calculate some of those two. What we're gonna do is just get started. We're gonna calculate the work done to the B by the isobaric compression first. Then we're gonna come back and calculate the work done due to the isothermal expansion. All right. So the isobaric compression, they recall that isobaric means that we have a constant pressure. And in an isobaric compression, the work done which we're calling WC here is just gonna be equal to the pressure P multiplied by the change in volume V two minus V one. Now, do we have all the information we need? Well, let's go ahead and write out the variables we have. So P were given in the problem and we're told that the pressure is 2.2 atmospheres, we wanna convert this to our standard unit. So what we're gonna do is we're gonna multiply by 1.013. Let me do this underneath. So we have room. So we're multiplying by 1.013 multiplied by 10 to the exponent five to convert to Pasco. Then we have our volumes. So V two is gonna be the final volume and V one is the initial volume. We're told that we are compressing from 5 L. That means V one is going to be 5 L 21.5 L. And so V two is 1.5 L. And again, we wanna convert these to our standard units. And so we're gonna multiply by 10 to the exponent negative three to convert to meters cubed. So five multiplied by 10 to the exponent negative 3 m cubed and 1.5 multiplied by 10 to the exponent negative 3 m cubed. So we have everything we need PV one and V two. We can go ahead substituting these values to calculate the first part of this work. OK. The work done in the compression. So WC is going to be equal to 2.2 multiplied by one point 013, multiplied by 10 to the exponent five pascals multiplied by 1.5 multiplied by 10 to the exponent negative 3 m cubed minus five multiplied by 10 to the exponent negative 3 m cu When we work all of this out, we get that this work done due to the Isobar compression is gonna be negative 780 501 Jules. All right. So we're gonna put a red box around this. So we don't lose track of it. We are going to need it to do that final sum of the works. OK? So we have the first part. Now we're gonna move to the work due to the expansion. So now we're looking at the isothermal expansion and recall that isothermal means that we have constant temperature. Now for an isothermal process, we're called the work done, which we're calling we due to work from the expansion is gonna be equal to N multiplied by R multiplied by T multiplied by the natural log. We're gonna call lawn of V two divided by the one. OK? All right. So what about the variables here? What do we have, OK. And you'll notice I didn't write all of the variables at the beginning. I've written them specifically for each process. And that's a really good idea in problems like this because the V one V two is P one P two is they change depending on which process you're looking at. Right. One is compressing, one is expanding, there is a difference. So in this case, we need N and we're told the number of moles in the problem is 0.32. So we have 0.32 mole, the gas constant are we know we don't know the temperature tea. We haven't been given that V two in this case. And V one are actually gonna be the opposite of what they were in the first case. OK. In the first one, we were starting with 5 L and compressing down to 1.5 L. In this case, we're starting with that 1.5 L and expanding to 5 L. And so V two is going to be 5 L. V one is going to be 1.5 L. And again, we can convert those the same as we did before 2 m cubed. So multiplying by 10 to the negative three. All right. What other values do we need? Well, right now, nothing in particular, but we do need the temperature T. All right. How can we find the temperature? T Well, recall that we can use our ideal gas law. In order to use our ideal gas law, we're gonna need to know the pressure. So let's just go ahead and write the pressure as well. And the initial pressure during our isothermal expansion is going to be the pressure at the end of our isobaric compression. Now remember that isobaric means we have constant pressure. So the pressure at the end is the exact same as the pressure that we started with, which was given in the problem 2.2 atmospheres. And we've already converted this. So 2.2 multiplied by 1.013 multiplied by 10 to the exponent five pascals. All right. So we have everything we need to calculate the temperature. T. So let's go ahead and find tea now and recall the ideal gas law, P one V one is equal to N RT one. OK. We're using this first point at the beginning of our isothermal expansion because we don't know the pressure at the end yet. So we're gonna use the, that initial time point. All right. So our temperature T one is therefore going to be equal to P one V one divided by Nr and I'm gonna replace this T one. I'm just gonna call it T and this is an isothermal expansion. So the temperature is the same throughout the entire process. So we're just gonna call it T OK. And that's gonna be the pressure 2.2 multiplied by 1.013 multiplied by 10 to the exponent five macos multiplied by the initial volume. 1.5 multiplied by 10 to the exponent negative 3 m scooped, divided by N 0.32 mole multiplied by R 8.314 D thermal Alvin. OK. Oh Now the unit of mole is gonna divide out the denominator. We have pascal meters cubed which is gonna be equivalent to that unit of jewel. So those are gonna divide out and we're gonna be left with just that unit of Kelvin in the numerator, which is what we want for temperature. And we get that the temperature is 125 0.65 Kelvin approximately. OK. So we have our temperature. That was the last thing we needed in order to calculate the work done in this isothermal expansion, we can get back to our we equation and we have that we is equal to N which is 0.32 mole multiplied by R 8.314 Joel per mole Calvin multiplied by the temperature 125.65. Kelvin multiplied by the natural log V two divided by V one. OK. So V 25 multiplied by 10 to the exponent negative 3 m cubed divided by 1.5 multiplied by 10 to the exponent negative 3 m skin. All right. Now, inside of this log the units are going to divide out. We're gonna have just a number. Take the log, we're just getting a number here. OK? No units. The rest of the equation we have moles and Calvin and then we're dividing by mole Kelvin. OK? So those are going to all divide it. We're gonna be left with just the unit of jewel, which is the unit we want for work. So that's great. And when we work this all out, we get the, the work done due to the isothermal expansion, it's gonna be 402 0.47 524 jewels. And again, putting a red box around that. So we don't lose track of it. That is half of our answer. OK? Now let's remind ourselves of what we're doing. We wanted to calculate this thumb of the works. OK? So WC plus we, so let's go ahead. Final step. We gotta add these two values together to get our results. So the work W is equal to WC plus WEKWC was that value we calculated in blue from the isobaric compression. We in green from the isothermal expansion you get that W is equal to negative 780.01 Joel. Yes, 402 547524 jewels. And when we work this all out, we get a final work of negative 377 0.53476 jewels approximately. And that is the sum of works that we were looking for. If we compare this to our answer choices and we round to the nearest jewel. We can see that the correct answer is going to be option C negative 378 jewels. Thanks everyone for watching. I hope this video helped you in the next one.

0.25 mol of a gas are compressed at a constant pressure of 250 kPa from 6000 cm^3 to 2000 cm^3, then expanded at a constant temperature back to 6000 cm^3. What is the net work done on the gas?

Verified Solution

Key Concepts

Work Done on a Gas

Ideal Gas Law

Isothermal Process