Two cylinders each contain 0.10 mol of a diatomic gas at 300 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.

a. What are the final temperature and volume of each?

Question

Two cylinders each contain 0.10 mol of a diatomic gas at 300 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.

a. What are the final temperature and volume of each?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Imagine two vessels that contain 0.20 moles of a mono atomic gas that are at identical initial conditions of 400 Kelvin and five atmospheres. Pressure vessel a undergoes an isothermal expansion while vessel b undergoes an adiabatic expansion until they both reach a final pressure of three atmospheres. What will be the final temperatures and volumes of this gas inside of each vessel? So that's our angle. Our angles are trying to figure out what the final temperatures, temperature values and volume values are for this particular gas inside of each vessel. So essentially we're trying to figure out what the temperature is and volume is for vessel A and B. So with that in mind, now that we know that we're solving for the temperature value, the final, specifically the final temperature value and the final volume value for vessel A and vessel B. Let's read off our multiple choice sensors to see what our final answer set might be. And let us note that all the temperature values are in units of Kelvin. And all the volume values are in units of cubic meters. And it's gonna go for vessel A first and then vessel B so A is 491 and 2.2 multiplied by 10 to the power of three for A and for B vessel deep, it's 326 and 2.8 multiplied by 10 to the power of negative three. For B for A, it's 402.2 multiplied by 10 to the power negative three. And for B it's 326 and 1.8 multiplied by 10 to the power negative three for C for A. It's 325 and 2.5 multiplied by 10 to the power of negative three. For B. It's 401.8 multiplied by 10 to the power of negative three. For D for A. It's 325 and 2.5 multiplied by 10 to the power of negative three. And for B it's 403.7 multiplied by 10 to the power of negative three. OK. So first off, let us investigate vessel A first which vessel A we're dealing with an isothermal expansion. So let us note that for this case and let's put a V dot A. So we know that we're talking about vessel A. So let us note that we have the initial conditions for vessel A T I, which is our initial initial temperature for vessel A is 400 Kelvin. And we also know that the initial pressure, which is P I is equal to five atmospheres, which P I, as we should note, I'm just gonna go ahead and give you the conversion, but you can use dimensional analysis or you quickly look this up on your own. But five atmospheres is also equal to 506625 pass scales. So it's the equivalent to 506,625 pass scales. So that's the same as five atmospheres. We also know the final pressure, which is PF and we know that PF is equal to three atmospheres, which will mean that PF our final pressure is also equal to 303,975 ass Awesome. So we need to note that this is an isothermal process. So make a little quick note here in blue, once again, this is important. So since this is an isothermal expansion, we should recall and remember that the final temperature will be equal to the initial temperature, which means that it will be equal to 400 Kelvin in this case, in this particular case. So with that in mind, moving along here, we now need to recall and use the ideal gas law, which as we should recall. The ideal gas law says that the pressure multiplied by the volume is equal to the number of moles multiplied by the universal gas constant multiplied by the temperature. So we need to rearrange this equation to isolate and solve for V. But it's very important that we solve for, we have to solve for the initial volume first in order for us to solve for our final volume value, which is what we're ultimately trying to solve for. So when we rearrange our ideal gas equation to isolate and solve for V I, in this case, since we're trying to solve for the initial volume, first, we will find that V I is equal to. And once again, we're just rearranging the ideal gas light equation to isolate and solve for V all by itself which will give us V I is equal to N multiplied by R multiplied by T. But specifically, since we're solving for the initial volume, we need to note that we're dealing with the initial temperature divided by the initial pressure. So we need to use the initial pressure value and the initial temperature value. So now at this point, all we have to do is just plug in all of our known values to solve for V I. So we know that N our number of moles is equal to 0.20 moles. And we need to multiply this by the universal gas constant, which we should recall this constant to be 8.314. And its units are Jews per Kelvin multiplied by mole. And we need to multiply this by our initial temperature, which we should recall is 400 Kelvin. Fantastic. And we need to divide everything by our initial pressure which we should recall from above that. It's 506,000 625 scales. Fantastic. So let me plug that into our calculator. We will find that V I writing it in scientific notation and rounding the two decimal places, you will get 1.31 multiplied by 10 to the power of negative three. And of course, its units will be cubic meters fantastic. So moving right along here, we can now solve for our final volume. But we must first recall that our final volume, sorry define the following volume. We need to recall and use the following equation which states that the final pressure multiplied by the final volume is equal to the initial pressure multiplied by the initial volume. So we need to rearrange this equation to isolate and solve for VF which is our final answer that we're ultimately trying to solve for is we're trying to figure out what the final volume is for vessel A and this is for vessel A. We also have to do this again for vessel B. So we will determine that the A F is equal to P I multiplied by V I divided by PF So now all we have to do is just plug in our known variables to solve for BF. So we know that P I is equal to 506,625 us girls. And we need to multiply this by our initial volume which we just determined to be 1.31 multiplied by 10 to the power of negative three cubic meters fantastic. And then we need to divide this by our final pressure, which we should recall our final pressure to be 303,000 975 pascals. Fantastic. So with that in mind, when we plug this into our calculator, we will find that VF writing it in scientific notation and rounding to one decimal place because all of our multiple choice answers are rounded to one decimal place. We will get 2.2 multiplied by 10 to the power of negative three me cubic meters. And that's it. That is our first answer for the volume for vessel A Awesome. And then we also know that the temperature. So I'll just say T for temperature is also equal to 400 Kelvin. And this is once again for vessel A Awesome. So now we could start solving for vessel B, which is our next step. So I'm gonna say V dot B for vessel B. So now we're solving for vessel B. So now we need to note that we're dealing with an adiabatic expansion for vessel B. So that would mean that our initial conditions once again, for this case, since they're both the same should be 400 Calvin. And our initial pressure once again is P I, which is equal to five atmospheres, which once again, we're just gonna skip. Since we're using past scale units, you need to note that it's 506 1006 125 pascals. And we also need to note that our final pressure once again is equal to 303. So 303,000 975 pascals. So in this case, since it's an adiabatic process, we need to figure out what the final temperature is because we do not know what the final temperature is. So we need to recall and use the equation to help us solve for the final temperature for an adiabatic process which states that the final temperature is equal to the initial temperature multiplied by the final, sorry, the initial pressure, the initial pressure divided by the final pressure all to the power of one minus gamma divided by gamma. OK. So now we also need to note that gamma, in this case will be equal to CP divided by CV, which is equal to five third where CP is the specific heat for the pressure. And CP is a specific heat for the volume which in this case, this expression will be equal to five thirds because we're dealing with an abatic process and we're dealing with a mono atomic gas. So with that in mind moving right along here, we can now plug in all of our known variables to solve for TF. So TF is equal to 400 Kelvin. Since once again, we're just plugging in all of our known variables to solve for TF multiplied by 506,625 skills divided by 303 1009 175. So 303,000 975 past scales which I'm gonna keep this simple. And instead of writing out the whole gamma expression for the power, when you simplify it, you will get negative two FS. So when you plug in five thirds into our power expression, you'll end up getting negative 2/5 as your new power value. So when you plug that into your calculator and you round to the nearest full number, you'll get 326 Kelvin. Fantastic. So now we need to you once again, recall and use the ideal gas law to solve for VF. So as we should know, note that VF is equal to N multiplied by R multiplied by TF. And note that we have to use the final, since we're trying to solve for the final volume, we're using the final temperature value divided by our final pressure value. And don't forget that N in this case is the number of moles and, and so N is the number of moles and then capital RS are universal gas constant. Once again. So plugging in our known values to solve for our final answer, we will get 0.20 moles means that's our number of moles multiplied by our universal gas constant, which is 8.314 and its units are once again jewels huh Calvin multiplied by more. And then we need to multiply this by our final temperature, which we should take a moment really quick to box because that's important because that's one of our values that we're trying to solve for. So we need to multiply this by 326 Calvin. Awesome. And we needed to divide everything by our final pressure which is given to us as 303,000 975 past scales. OK. Fantastic. So let me plug that into our calculator. We will find that VF is equal to rounding to one decimal place to match all of our multiple choice answers 1.8 multiplied by 10 to the power of negative three. And once again, our units will be cubic meters and that's it. We've officially solved for this problem. Hooray, we did it. So let's look at our multiple choice answers to see which answer corresponds with the answers that we found together. So we know that vessel A was equal to 400 °C. And then we know like as we should recall, that are volume was 2.2 multiplied by 10 to the power negative three. So that means the final answer has to be the letter B which means that vessel A has a temperature of 400 Kelvin. And the volume is equal to 2.2 multiplied by 10 to the power negative three cubic meters. And vessel B's temperature is 326 Kelvin and the volume is 1.8 multiplied by 10 to the power of negative three cubic meters. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Two cylinders each contain 0.10 mol of a diatomic gas at 300 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.

a. What are the final temperature and volume of each?

Verified Solution

Key Concepts

Isothermal Process

Adiabatic Process

Ideal Gas Law

Two cylinders each contain 0.10 mol of a diatomic gas at 300 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.a. What are the final temperature and volume of each?

Verified Solution

Key Concepts

Isothermal Process

Adiabatic Process

Ideal Gas Law

Two cylinders each contain 0.10 mol of a diatomic gas at 300 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.

a. What are the final temperature and volume of each?