Skip to main content
Ch 19: Work, Heat, and the First Law of Thermodynamics

Chapter 19, Problem 19

A gas cylinder holds 0.10 mol of O₂ at 150°C and a pressure of 3.0 atm. The gas expands adiabatically until the volume is doubled. What are the final (a) pressure and

Verified Solution
Video duration:
0m:0s
This video solution was recommended by our tutors as helpful for the problem above.
105
views
Was this helpful?

Video transcript

Hey, everyone in this problem, we're told that at an initial pressure of four atmospheres and a temperature of 200 °C, 0.15 mole of N two gas is present in a closed vessel following an adiabatic expansion process, the volume of the gas triples. And we're asked to determine the final pressure of the nitrogen gas within that vessel. We have four answer choices all in pascals option A 6.25 times 10 to the exponent three. Option B 3.41 times 10 to the exponent four. Option C 9.94 times 10 to the exponent three. And option D 8.70 times 10 to the exponent four. So let's start by writing out some information we have and we're gonna use a subscript one to indicate that initial situation that we have. And we're gonna use subscript two to be the final situation. OK. So our initial pressure P one we're told is four atmospheres. Yes, we have four atmospheres. Now we're doing our calculations, we wanna be in our standard unit. We're gonna want to convert this to Pascal. And so what we're gonna do is take our four atmospheres, we're gonna multiply by 1.01 three times 10 to the exponent five to convert to pascals. Now, for our temperature, our initial temperature we're told is 200 °C. Again, we want this in our standard unit. So we're gonna convert to Calvin to do that. We're gonna add 273 when we get our temperature of 473 Kelvin, they were told that we have 0.15 moles. So N is gonna be 0.15 mole and we have our volume V one and we aren't given a numerical value for that. So we're just gonna call that V for right now. Now, in this final situation, OK, we're looking for our final pressure, which we're gonna call P two, the temperature T two. We don't know anything about the number of mole N is gonna be the same in our volume V two. While we're told that the volume triples, so V two is gonna be equal to three multiplied by V that initial volume. All right. So let's get started. We have some information about volume. We wanna find the pressure. They were told this is an antibiotic process. So we're called that we have the following equation V one multiplied by V one to the exponent gamma is equal to P two multiplied by V two to the exponent gamma. We have P one. We wanna find P two. So we don't know the values for V one and V two, but we do have a relationship between the two. So we're gonna start with that. We're gonna see if we can work through this without knowing those exact values. And the other thing we need here is gamma. Now we're dealing with N two. OK? We wanna find gamma. We're dealing with N two which is a diatomic gas. Yeah, because it's diatomic that tells us, let's see. B it is going to be, but pas are, we're assuming that N two is a nice ideal gas tells us that CP is going to be five halves R plus R. It's always CV plus R for an ideal gas which gives us seven halves R. And so what we get is like gamma, which is the ratio of the two CP divided by CV is going to be seven fifths and that's gonna be the same gamma for any diatomic ideal gas. So we have gamma. Now let's get back to our equation and we're gonna substitute in some numbers. So P one, we have is four multiplied by 1.013 multiplied by 10 to the exponent five. Let go the one which we're just calling the to the exponent gamma which is seven fifths or we can write this as a decimal of 1.4. So we have to the exponent 1.4. This is gonna equal P two, the value. We are trying to find that final pressure multiplied by V two, which we know is going to be three V and that's all raised to that same exponent of 1.4. OK. So let's expand a little bit on the right hand side here and see if we can do something with these V terms. So we're gonna leave the pressure alone four multiplied by 1.013 multiplied by 10 to the exponent five pa scales multiplied by V to the exponent 1.4 that's equal to P two. And we can apply this exponent to each of our terms. So we get three to the exponent 1.4 multiplied by V to the exponent 1.4. Let's give ourselves some more room to work. And what we can see now is that these V to the exponent 1.4 we can divide it and we divide both sides by V to the exponent 1.4. That term completely goes away and we're left with just P two. As the only unknown. We get that P two. It's gonna be equal to four, multiplied by 1.013 multiplied by 10 to the exponent five scales divided by three to the exponent 1.4. We work this out on our calculators. We get that, that final pressure P two, about 87,000 and 36.2. Pasco our answer choices had this with um three significant digits, we wanna convert this to scientific notation. And so we are going to do just that and we can write this as 8.70 multiplied by 10 to the exponent for high scales. And that is the final pressure that we were looking for. If we compare this to our answer choices, we can see that this corresponds with answer choice D. Thanks everyone for watching. I hope this video helped you in the next one.