The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mol of an ideal gas.
(a) What was the lowest temperature the gas reached in this process? Where did it occur?

Question

The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mol of an ideal gas.

pV-diagram showing process abc for 0.0175 mol of ideal gas in thermodynamics.

(a) What was the lowest temperature the gas reached in this process? Where did it occur?

pV-diagram illustrating process abc with points a, b, and c for an ideal gas.

Accepted Answer

Everyone in this problem. We have 0.023 moles of an ideal gas. That follows the process in the Diagram marked by JKL. And were asked to determine the least temperature of the gas and the point where it occurs. Alright, so let's look at this diagram. What we'll see is that this is a pressure volume diagram. K P V. We're asked to find information about temperature. Okay, So we want to relate pressure, volume and temperature. Now we have an ideal gas. Okay, Which means that we can use our ideal gas. The ideal gas law recall tells us that the pressure times the volume PV is equal to end. The number of moles are the gas constant times the temperature. Alright, so we have three points to consider. Okay, we have the point jay, we have the point K. And we have the point. Okay, so let's go ahead and find the temperature at each point using our ideal gas law and then we can compare them to figure out which one is the least. So let's go start with at J. Okay, so we have PV is equal to N R. T. Okay, we're gonna get the value for P and V from the graph, from the diagram. Okay. And that J we see that we are at a V a value of one leader. Okay. and AP of 0.5. S R P is going to be 0.5. This is a T. M. So we want to multiply by 1.13 to 5 times 10 to the five to convert to pascal's. And then we have our volume which is one liter and we want to convert to meters cubed. And so we have one times 10 to the negative three m cubed. And this is going to equal to end the number of moles. 0.023 moles. Okay. Times are the gas constant? 8.314 K. Times the temperature T. The unit of this gas gas constant we missed here is jules per mole Calvin. Okay we'll just write it in there so we don't forget. So on the left hand side we have Pascal's times meters cubed. Well that's equal to a jewel. So we get 50. jewels. Okay and on the right hand side we have moles times jewel per mole Calvin. Okay so we're gonna be left with 0. one two jules per Calvin. Okay. Times the temperature. T. And again the temperature T. Is what we are looking for here. Okay And we get the temperature T. Is equal to when we divide jewel by jewel per Calvin. We're gonna be left with just Calvin. The unit we want for temperature. So we get 264.9 for Calvin. Okay so the temperature at Point J. 264.94 Calvin. Let's move on to Point Kay. Okay and again ideal gas law PV. Is equal to N. R. T. Let's give ourselves some more room. The pressure P at point K. Is 0.25 A. T. M. And the volume is three L. Okay. And so we're going to get 0. Times 1.013-5 times 10 to the 5K to convert those atmospheres into Pascal's the volume three liters. Okay. Times 10 to the negative three to convert to meters cubed. Okay. And on the right hand side N. R. Is the same. Okay, no matter what point we're at we'll have the same number of moles in the gas constant is a constant. And so we're just gonna have 0.191 to 2 to jules per Calvin times the temperature T. Here as well. Okay, so that saves us writing that out On the left hand side. We get 75. jewels Is equal to 0.191222 Jules per Calvin T. Okay. And when we divide the unit of jewel will cancel. And we're left with the temperature t of 397. to Calvin. All right. And we'll see here that the temperature is bigger than the temperature at J. Okay, so so far J is the smallest temperature. We just have to look at point L one thing I want to note here now that we've seen this twice this N. R. On the right hand side. That the constant we used. Okay, what that tells us is that whichever point has the largest value for PV is going to have the largest value for the temperature. T Okay so we could just find PV for each of these, compare them and then find the temperature for one. I'm going to continue like this so we can see the full process. Okay and this time we're going to point L. Let's just before we scroll down to have more room to work. Let's just look at L on our diagram and pick out the pressure and volume. Okay L is going to have a pressure of 0.75 A. T. M. And a volume of three L. Alright, so growing down accidentally deleted our unit kelvin there. Okay so I point out again the ideal gas law PV is equal to N. R. T. Our pressure we found from our graph. 0.75. Okay, it's an atmosphere. So we multiplied by 1.013-5 times 10 to the five. To turn that into Pascal's hey, times are volume which is three leaders. Okay, so three times 10 to the negative three to convert to meters. Cute. And on the right hand side again we have that same and our constant 0.191222 jewels per Calvin times the temperature T the left hand side. Multiplying we get 227. pascal meters cube gives a unit of jewels. The right hand side is unchanged. 0.191 to 2 to jules per Calvin times. T. Okay. When we divide jewels by Jules for Calvin we get a unit of Calvin, which is what we want for temperature we get a temperature 1,192. Calvin. Okay, so again this is a bigger temperature. And so if we compare our temperatures, the smallest one we got was the temperature 264.94. This was the smallest temperature we got. And it occurred at point J. So our answer is going to be e smallest temperature is 265 Calvin at Point J. Thanks everyone for watching. I hope this video helped see you in the next one.

The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mol of an ideal gas. (a) What was the lowest temperature the gas reached in this process? Where did it occur?

Verified Solution

The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mol of an ideal gas.
(a) What was the lowest temperature the gas reached in this process? Where did it occur?