(III) One mole of monatomic gas undergoes a Carnot cycle with ...

Question

(III) One mole of monatomic gas undergoes a Carnot cycle with T_H = 350°C and T_L = 210°C. The initial pressure is 8.8 atm. During the isothermal expansion, the volume doubles. Determine Q, W, and ∆E_int for each segment of the cycle.

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. A heat engine operates using one mole of a mo atomic gas in a thermodynamic cycle. The engine absorbs heat at 500 Kelvin and releases it at 300 Kelvin starting from state one, the gas has a pressure of 2.5 atmospheres and a volume of 16.4 L. During isothermal expansion to state two, the volume doubles to 32.8 L and the pressure drops to 1.25 atmospheres. During an adiabatic expansion to state three, the volume increases to 66.7 L and the pressure drops to 0.37 atmospheres, isothermal compression to state four reduces the volume to 33.4 L and the pressure increases to 0.74 atmospheres. Adiabatic compression returns the system to state one, calculate the heat transferred Q work done W and change in the internal energy delta E eternal for each segment 1 to 22 to 3 processes only. OK. So it appears for this particular prong we're solving for three separate values we're solving for the heat transferred Q, the work done W and the change in internal energy delta E eternal or two separate processes. So we're trying to solve for these three values. So we're trying to solve for QW and delta E eternal for segment 1 to 2 and 2 to 3. So two separate processes. Awesome. So with that in mind, let's quickly look at our diagram that's provided to us by the prom itself before we look at our multiple choice answers. And as you can see, we have a standard PV diagram, we have our pressure denoting our Y axis versus our volume which is the X axis. And in green, we have our processes denoted by bolded green arrow showing the flow or the transition. So from state one to state two, as we discussed is an isothermal expansion. And from state 2 to 3, we have an adiabatic expansion and then from state 3 to 4, we have an isothermal or an isotherm compression. And then from states 4 to 1, we have an adiabatic compression. So from states 3 to 4 and 4 to 1 are compression states, whereas 1 to 2 and 2 to 3 are expansion states. Awesome. And as we could see, we have our two different temperatures. So we know that the engine absorbs heat at 500 Kelvin and releases it at 300 Kelvin. And our dotted black lines denote our temperatures at, for 300 Kelvin and at 500 Kelvin. OK. So now that we have a visualization of what's going on in this particular pro let's read off our multiple choice answers to see what our final answer sets might be noting that we're gonna read them in order from process 1 to 2 1st and then secondly, process 2 to 3. And the order of our variables will be delta E eternal first. Then our value for Q then our value for W so A is zero and 2.9 multiplied by 10 to the power of three and 2.9 multiplied by 10 to the power of three. And I forgot to mention that all of our values for delta E eternal for Q and for W are all in the same units of jewels. And for our second process, it's negative 2.5 multiplied by 10 to the power of 30 and 2.5 multiplied by 10 to the power of three Jews. OK? And then B is zero and 3.0 multiplied by 10 to the power of three and 3.0 multiplied by 10 to the power of three. And for a second process, 2.5 multiplied by 10 to the power of 30 and negative 2.5 multiplied by 10 to the power of three C is 02.9. Multiplied by 10 to the power of 32.9 multiplied by 10 to the power of three. And for our second process negative 2.0 multiplied by 10 to the power of 30 and positive 2.0 multiplied by 10 to the power of three. And finally D is 03.0 multiplied by 10 to the power of three and 3.0 multiplied by 10 to the power of three. And our second process is negative 2.0 multiplied by 10 to the power of 30 and 2.0 multiplied by 10 to the power of three. OK. So now that we have an idea of what our possible final answer set might be. Let's start to solve our problem. So first off, in order to help us solve this problem, let's write down all of our given variables, all of our known variables. We know the hot temperature which is equal to 500 Kelvin, which we'll call this th th is equal to 500 Kelvin. We also know our, I said hot temperature, but let's say high. So we know our high temperature is 500 Kelvin. And we also know our low temperature which we'll call TL and our low temperature is equal to 300 Kelvin. Awesome. We also know our P one value, our pressure one value and our pressure one value is equal to 2.5. And don't forget that its units are atmospheres. We also know our volume one or V one value and this is equal to 16.4 L. Fantastic. That is our first volume. That's our V one value once again. So it's 16.4 L. We also know our volume to value and our volume two value. So this is where the volume doubles during isothermal expansion. And this is equal to 32.8 L. Fantastic. We also know our P two value or second pressure value which is equal to 1.25 atmospheres. And we also know our V three, our third volume which is equal to 66.7 L. And we also know our final pressure. So P three and our P three pressure or third pressure is equal to 0.37. And of course, its units are atmospheres awesome. So let us note, we are now going to have to start solving for QW and delta E eternal for each segment of the cycle. So let's focus first on our isothermal process, which we're gonna be focusing on the segment 1 to 2. And when we focus on 1 to 2, we need to note that delta E eternal will be equal to zero. So that is one of our final answers. So let's put that in a box since that's important to know. Awesome and moving right along here. And our next step is we need to start solving for Q and W So this is the work done in the heat transferred. So let us note that Q 12 is equal to W 12, which is equal to when we plug in our known variables, 1.0 moles visits the amount of our sample that we are given multiplied by our universal gas constant, which is 8.314. And don't forget that its units are joules per mole multiplied by Kelvin. And then we need to multiply it by our high temperature which is 500 Kelvin. And then we need to multiply it by our natural log of 32.8 L divided by 16.4 L. Fantastic. And then when we plug that into our calculator, we will find that this is equal to 2881.41 jules, which is approximately equal to when we round to one decimal place. And we write it in scientific notation, 2.9 multiplied by 10 to the power of three jewels. And that is our final answer for Q and for W. So this is equal to Q and it's also equal to W Awesome. So now we can start solving for our second process which is 2 to 3 and focusing on 2 to 3. Right now, we need to note that in this particular case, Q 23 is equal to zero. And let's quickly note that we're just so we don't get confused that each time we say zero to be more specific, we're saying zero jewels awesome and moving right along here. So now we need to solve for delta E eternal. So delta E eternal is equal to three halves. So three halves multiplied by lowercase N multiplied by capital R multiplied by T three minus T two, which is all equal to three halves. So three halves multiplied by when this is now when we're plugging in our known variables, 1.0 moles. So once again, we're taking our equation and we're plugging in all of our known variables. And then we need to multiply it by R which R is our universal gas constant. Once again, lowercase N is our number of moles. So once again, our universal gas constant is 8.314. And don't forget that its units are joules per mole multiplied by Kelvin. Fantastic. So now we need to take our universal gas constant and we need to multiply it by our temperature which is T three minus T two. So we need to take 300 Kelvin. And ultimately, we're getting all this information from the figure or the diagram that's provided to us by the prom itself. So there's no room for confusion here. Awesome. So it's 300 Kelvin minus 500 Kelvin. So let me plug that into our calculators. We'll get negative 2494.2 jewels, which is approximately equal to when we round to one decimal place, when we round it, rounding the one decimal place in scientific notation, which will give us negative 2.5 multiplied by 10 to the power of three tools. So that is our final answer for delta E eternal. So this is equal to delta E eternal. Awesome. So now we need to quickly note here, this is very important that W-2 3. So W-2 3 is equal to negative delta E eternal. So that means that our final answer for W will be equal to positive 2.5 multiplied by 10 to the power of three. Because initially our Q or for I rather our delta E eternal value is equal to negative 2.5 multiplied by 10 to power three. But a negative multiplied by a negative gives us a positive. Awesome. So that is our final answer for W hooray. We did it. So looking at our multiple choice answers, the correct answer has to be multiple choice answer letter A which once again is process 1 to 2 delta E eternal is equal to zero. Joules Q is equal to 2.9 multiplied by 10 to the power of three. Juls W is equal to 2.9 multiplied by 10 to the power of three Joules process 2 to 3. Delta E eternal is equal to negative 2.5 multiplied by 10 to the power of three. Joules Q is equal to zero which I forgot to box that. But let's box that really quick right now. So Q is equal to zero jewels and W is equal to 2.5 multiplied by 10 to the power of three jewels and that's it we've officially solved for this problem. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

(III) One mole of monatomic gas undergoes a Carnot cycle with T_H = 350°C and T_L = 210°C. The initial pressure is 8.8 atm. During the isothermal expansion, the volume doubles. Determine Q, W, and ∆E_int for each segment of the cycle.

Key Concepts

Carnot Cycle

First Law of Thermodynamics

Isothermal Process

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(III) One mole of monatomic gas undergoes a Carnot cycle with TH = 350°C and TL = 210°C. The initial pressure is 8.8 atm. During the isothermal expansion, the volume doubles. Determine Q, W, and ∆Eint for each segment of the cycle.

Key Concepts

Carnot Cycle

First Law of Thermodynamics

Isothermal Process

Watch next

(III) One mole of monatomic gas undergoes a Carnot cycle with T_H = 350°C and T_L = 210°C. The initial pressure is 8.8 atm. During the isothermal expansion, the volume doubles. Determine Q, W, and ∆E_int for each segment of the cycle.