(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.

(a) Find the values of the pressure and volume at the points a, b, c, and d of Fig. 20–5.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the car, no cycle. This problem says picture a single mole of a diatomic gas undergoing various stages of a car. No cycle. As shown in the figure is initially at a temperature of 400 °C. The pressure experienced by the gas at this stage is 10 atmospheres which is point A on the diagram at the end of the isothermal expansion, its volume triples at point B. Afterwards, the gas temperature falls to 250 °C at point C calculate the pressure volume at different points of the cycle points. ABC and D below the question, we were given a DI A PV diagram of the cardinal cycle that was described in the problem. Now, this question asks us to calculate the pressure and volume at the various points. So we're actually gonna start at point A because we have the most information about that point. Now we're told that its temperature which we'll call T A is equal to 400 °C. But I'm gonna need that in Kelvin. So I'm gonna take that 400 add 273 to it. And that'll give you my temperature in Kelvin. That means that T A is gonna be equal to 673. Kelvin, we're also told the pressure at this point. So P A is gonna be our pressure and that's gonna be equal to 10 atmospheres. And I think one of the quantities that I needed to find, the other quantity that I need to find is the volume which we'll call VA. And since I have the pressure and temperature, we can use the ideal gas law to calculate our volume. So recall from the ideal gasol, that VA is gonna be equal to N RT A divided by P A. We're in, here's the number of moles and Rs are gas constant. So we can go ahead and plug in value. So we'll have VA is equal to the number of moles is gonna be one since we told, we're told that we have a single mole of gas in the problem. So we'll have one mole multiplied by our gas constant, which has a value of 0.0821 that has units of atmospheres liters per mo Kelvin, that's gonna be multiplied by our temperature T A which is 673 Kelvin. And this is why we had to convert our temperature into Kelvin to cancel the units of Kelvin in our gas constant. Now, that whole quantity now is gonna be divided by our pressure which is 10 atmospheres. So plugging those values into our calculator will get va is equal to 5.5 three leaders. But I went to round this to two significant figures to match the number of significant figures that I was given the problem. So this is gonna be approximately equal to 5.5 L. And so that is my volume that I needed to calculate it. So I now have both the pressure and volume calculated for point A. So we're just gonna move on out around the cycle and calculate the pressure, volume and temperature at each point. So we're gonna move on to point B. So at point B, since we're going from point A to point B through an isothermal process, that means the temperature remains constant. That means the TB or temperature at B, it's gonna be equal to T A which is gonna be equal to the 673 Kelvin. We're also told in the problem that our volume might be Phoebe is equal to three times that of VA we're told that the volume triples going from point A to point B. So we can actually plug in values. So we'll have VB is equal to three, multiplied by the volume that we calculated four point A which is the 5.53 L. So that means that VB gonna be equal to 16.6 L which again will round to two significant figures. So that's gonna be approximately 17 L. But now have the volume calculated at point B. And since I have the temperature and volume at point B I can use again the ideal gas law to calculate the pressure. So we'll have from the ideal gas law, we'll have P A or sorry, PB is equal to in RTB divided by VB again, plug in the value. So we'll have PB, it's equal to one more multiplied by 0.0821 atmosphere's liters per mole. Kelvin, that's multiplied by 673 Kelvin. And that whole quantity is divided by the volume that we just calculated, which is the 16.6 L. So plugging those into my calculator, I'll get PB is equal to 3.33 atmospheres. And again, we're just going to round that to two significant figures. So that's gonna be approximately 3.3 atmospheres. So that is the pressure that I was looking for for point B. So I now have both the volume and pressure at point B. So now we're going to move along to point C. Now to go from point B to point C, we're looking at an adiabatic process. However, we were actually told the temperature at point C and the problem. So TC is going to be equal to 250 °C. So we'll need to convert that into Kelvin. So I'll take the 250 add 273 to it to convert my temperature into Kelvin. That means TC is going to be equal to 523 Kelvin. Now, next we're gonna look at is the volume and we actually have a relationship between the volume and temperatures at the two endpoints for an adiabatic process. And so recall for that relationship, we have TC multiplied by VC raised to the gamma minus one is equal to TB multiplied by VB raised to the gamma minus one. And here gamma is the ratio of our specific heat capacities for the ideal gas at constant pressure and constant volume. Now, we actually want to solve this for VC. So doing so we'll get VC is equal to VB, what plane, the quantity of TB divided by TC in quantity and that's raised to the quantity of one divided by the quantity of GAMMA minus one. And so we actually have values for everything. On the right hand side, we'll plug them in. So we'll have VC is equal to or VB, that is the 16.6 L that could be multiplied by TB, which is the 673 Kelvin divided by TC, which is the 523 Kelvin. And that's gonna be raised to the quantity of one divided by the quantity. And here gamma um is going to be seven divided by five. That's because we're told that we have a diatomic gas in the problem. So that means gamma is seven divided by five for diatomic molecules. And then I'll have the seven divided by five minus one in the denominator of that exponent. So, plugging this into my calculator, I'll get VC is equal to 31.2 L. And again, I wanna approximate that as 31 L to round it to two significant figures. That is my volume at point C and now that I have both the volume of temperature, I can use those and the ideal gasol to calculate my pressure. So we'll have PC is equal to N RT C divided by VC and plugging in the values we get PC equal to one more multiplied by 0.0821 atmosphere's leaders per more. Kelvin, that's multiplied by our temperature TC, which is the 523 Kelvin and all that is going to be divided by VC, which is the 31.2 L. So plug in those values into our calculator, we'll get PC is equal to 11.38 atmospheres. And again, we'll need to round that to two significant figures. So it's gonna be approximately 1.4 atmospheres do now have both the pressure and volume for point C. So now we can move on to point D now for point D to go from point C to point D is an isothermal process. So that means that the temperature has to remain constant. That means that TD is equal to TC, which is equal to the 523 Kelvin. Now, in order to calculate the volume, we're actually going to look at the adiabatic process, go going from point to point A. So we're gonna use the same relationship between the temperature and volumes at those two points that we did in four point C. So here we're going to have PD is equal to va multiplying the quantity of T A divided by TD in quantity that's raised to the one divided by the quantity of gamma minus one. So we have qua uh values for all the quantiles on the right hand side, we can go ahead and plug those in. And so we'll have VD is equal to or VA that was the 5.53 L multiplying the quantity for T A. That is the 673 Kelvin divided by the TD, which is the 523 Kelvin. That quantity is gonna be raised to one divided by the quantity of seven, divided by five minus one. So plugging those values into our calculator will get VD is equal to 10.4 L. And again, we're gonna approximate that as 10 L to round it to two significant figures. That's my volume calculation. And again, I have the volume and temperature. So I can use the ideal gas law to calculate my pressure. So in this case, we'll have PD is equal to NR TD divided by VD and plug it in values will have PD is equal to one more multiplied by 0.0821 atmosphere's liters per more. Kelvin, that's multiplied by TD, which is 523 Kelvin. And that's gonna be divided by PD, which is the 10.4 L. So plug in those values to my calculator, I'll get PD is equal to 4.13 atmospheres. And again, we'll round that to two significant figures. That's gonna be approximately 4.1 atmospheres. So that is gonna be my pressure value at point D. So now we have actually calculated the volume and pressure at every single point. So just to recap real quick, what we've done, we calculated the pressure volume and temperature at every single point. And to do that, we actually utilize our relationship between those three quantities for both an idiotic and isothermal process along with our formula for the ideal gas law. And so that allowed us to calculate the pressure and volume at every single point, which is what the question was looking for. So I hope that this has been useful and I'll see you in the next video.

(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.

(a) Find the values of the pressure and volume at the points a, b, c, and d of Fig. 20–5.
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Key Concepts

Carnot Cycle

Ideal Gas Law

Isothermal Process

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