A reversible heat engine operates between two reservoirs, T1 and T2 (T2>T1) . T1 can be consider

A reversible heat engine operates between two reservoirs, T1...

A reversible heat engine operates between two reservoirs, $T_1$ and $T_2$ $\left(T_2>T_1\right) . T_1$ can be considered to have infinite mass, i.e., $T_1$ remains constant. However the warmer reservoir at $T_2$ consists of a finite amount of gas at constant volume ( $\mu$ moles with a specific heat capacity $C_v$ ). After the heat engine has operated for some long period of time, the temperature $T_2$ is lowered to $T_1$. (a) What is the heat extracted from the warmer reservoir during this period? (b) What is the change of entropy of the warmer reservoir during this period? (c) How much work did the engine do during this period?

A reversible heat engine operates between two reservoirs, $T_1$ and $T_2$ $\left(T_2>T_1\right) . T_1$ can be considered to have infinite mass, i.e., $T_1$ remains constant. However the warmer reservoir at $T_2$ consists of a finite amount of gas at constant volume ( $\mu$ moles with a specific heat capacity $C_v$ ).

After the heat engine has operated for some long period of time, the temperature $T_2$ is lowered to $T_1$.
(a) What is the heat extracted from the warmer reservoir during this period?
(b) What is the change of entropy of the warmer reservoir during this period?
(c) How much work did the engine do during this period?

Key Concepts

Work and Energy Conservation

Work in thermodynamics refers to the energy transferred when a system exerts a force over a distance. It is governed by the principle of energy conservation, where the work performed by a heat engine is equal to the net heat absorbed minus the heat rejected. This relationship is critical for understanding the efficiency and performance of the heat engine during the energy exchange processes.

Entropy Change

Entropy, a measure of disorder, quantifies how energy is distributed in a system. In thermodynamic processes, especially reversible ones, the change in entropy is calculated by integrating the heat transfer divided by the temperature (?S = ?dQ/T). This concept is fundamental in tracking how energy transfer affects system disorder and the feasibility of processes.

Specific Heat Capacity at Constant Volume

The specific heat capacity at constant volume determines the amount of heat required to raise the temperature of a substance by a certain amount while its volume remains fixed. This property is essential for quantifying the energy change in a finite reservoir, where the temperature variation is directly related to the heat extracted or added.

Thermal Reservoirs

Thermal reservoirs are bodies that can exchange heat with a system at constant temperature. In thermodynamic analysis, one reservoir is often assumed to be infinite, meaning its temperature remains constant during heat exchanges, while a finite reservoir has a limited heat capacity and its temperature changes as energy is transferred.

Reversible Heat Engines

Reversible heat engines operate in idealized processes where no entropy is produced, meaning the process can be reversed without net changes in the universe's entropy. They represent the maximum efficiency achievable under the laws of thermodynamics and are typically modeled by Carnot engines, which set the benchmark for efficiency between two temperatures.

Finite Versus Infinite Reservoirs

Understanding the difference between finite and infinite reservoirs is crucial. Infinite reservoirs maintain a constant temperature regardless of the heat exchanged, whereas finite reservoirs, due to their limited heat capacity, will change temperature during the process, affecting both the amount of heat transferred and the entropy of the system.

Transcript

00:02 So we are asked to calculate the change in entropy of the cold reservoir during the step b process.

00:10 Now before we go ahead and do that, i just want to say that in order to tackle this problem, it is very useful to first sketch it out on a pv plot, on a pressure volume plot that is.

00:21 And so what we have here is that during process a, we have an isothermic process for the gas, traversing this green curve, where the temperature of the isotherm is 373 degrees kelvin.

00:35 And then we have an isochoric process, isochoric meaning constant volume, where the process travels from p2 v2 to p3v2.

00:46 And then we have another isothermic process at 273 kelvin, traveling from v2, p3 to v1, p4.

00:55 And then we have finally another isochoric process where the gas travels from p4 v1 back to p1 v1.

01:05 And so it's important to note that during the two isochoric processes in blue, no work is done on or by the gas because by the very definition of work, as the volume under the curve on a pv plot, there is no, sorry, as the area under the curve on a pv plot, there is no area.

01:29 Since it is a straight vertical line at constant volume.

01:34 And so there's no pressure volume work done during those two periods.

01:38 And so if we now go ahead and calculate the entropy change during part b, by the change in entropy formula that we are familiar with, change in entropy is equal to the amount of heat flow over the constant temperature.

01:55 We know that during the process b, as we just argued, that by the first law, delta u is equal to q plus w, but w is zero since there is no area underneath the the trajectory in the process labeled by b and so delta u is now only equal to q but q we know is equal to n c u delta t that is the amount of heat flow into or out of the gas is is equal to n where n is the number of moles and the heat capacity at constant volume because this is a constant volume process multiplied by the change in temperature and so we can go ahead and calculate this directly so we are calculating the change in internal energy of the gas because we want to know how much heat has been removed from the reservoir and so when we calculate this we have that cv for a diatomic ideal gas is five halves r and so plugging this in we have n r and so plugging this in we have n 5 halves multiplied by delta t and now using the ideal gas law we can substitute out nr so we have that pv is equal to nrt and we solve for nr using some values of pv and t on the b curve and so i will choose to substitute out p2 v2 so p2 v2 over the temperature on the b trajectory initially multiply by a change in temperature.

03:47 And so just to explain this a little bit more, we have, i'm taking the point, i'm taking the point on the b trajectory at this point here, the red circle.

04:01 So it's at a pressure of p2, at a volume of v2, and its temperature at that point initially would simply be 373 kelvin because it is coming to a close of the isothermic trajectory where the temperature, of the gas was constant at 373 degrees kelvin.

04:22 And so now the change in temperature is, now we can simply plug in our values, and so we have that the amount of heat flow is equal to five halves multiplied by 1 .24 times 10 to the 5 pascales, multiplied by 2 .50 times 10 to the minus 2 meters cubed, multiplied by the change in temperature.

04:54 So we have 273 kelvin minus 373 kelvin all over 373 kelvin.

05:08 And so when we calculate this, we obtain negative 2077 .75 joules.

05:15 So this is the amount of heat that has traversed out of the gas.

05:22 And we know that it is out of the gas.

05:24 Because we have a negative sign here, which was taken care of by the change in temperature factor, where the initial and final temperature difference has given us a negative sign.

05:37 And so now we can calculate the change in entropy of the reservoir because we know that this amount of heat that has flown out of the gas has simply traveled into the reservoir.

05:51 And so the amount of heat into the reservoir is positive 20, 77 .75 joules over the constant temperature of the reservoir, which is 273 kelvin, a result of 7 .611 joules per kelvin.

06:11 This is the change in entropy of the reservoir at process b.

06:22 And so for the next part, we want to calculate the change in entropy of the reservoir in step d.

06:32 So in the isochoric process where the gas travels from p4v1 to p1v1.

06:39 And so we can do the same thing once again where we calculate the amount of heat flow out of the gas at that portion or into the gas and state that by conservation of energy, this is the same amount of heat that has moved in or out of the reservoir that it is in contact with and then calculate the change in entropy in that way...

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