The operation of a certain heat engine takes an ideal...

The operation of a certain heat engine takes an ideal monatomic gas through a cycle shown as the rectangle on the $P V$ diagram of Fig. $25 .(a)$ Determine the efficiency of this engine. Let $Q_{H}$ and $Q_{L}$ be the total heat input and total heat exhausted during one cvcle of this engine. (b) Compare (as a ratio) the efficiency of this engine to that of a Carnot engine operating between $T_{\mathrm{H}}$ and T_{\mathrm{L}},$ where $T_{\mathrm{H}}$ and $T_{\mathrm{L}}$ are the highest and lowest temperatures achieved.

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics, which is the principle of energy conservation, states that the change in internal energy of a system equals the net heat added to the system minus the work done by the system. This concept is key to analyzing the energy flow and determining both the work performed and the heat exchanged in thermodynamic cycles.

PV Diagram and Thermodynamic Cycles

A pressure-volume (PV) diagram is used to graphically represent the various processes in a thermodynamic cycle. By plotting pressure against volume, one can visualize the work done during expansion or compression and analyze different segments of the cycle, such as isochoric and isobaric processes, which are common in many engine cycles.

Heat Engine Efficiency

The efficiency of a heat engine is the ratio of the net work output to the total heat input. It is a measure of how effectively an engine converts absorbed heat into work, and it is fundamentally constrained by the second law of thermodynamics.

Carnot Cycle Efficiency

The Carnot cycle represents an idealized heat engine operating between two temperature reservoirs. Its efficiency, given by one minus the ratio of the lower temperature to the higher temperature (? = 1 - T_L/T_H), sets the theoretical upper limit on the efficiency of any real heat engine working between the same temperature limits.

Transcript

00:01 In this question, we have mono -atomic gas that is being used for certain heat engine that utilizes this cycle.

00:11 Okay, there are two parts in the question.

00:13 We want to find out the efficiency and compare this efficiency to the efficiency to the efficiency of the carnot engine.

00:25 Alright, so part a, k, note that efficiency, efficiency, okay is equal to w divided by qh okay so you need to identify which of the four steps you have qh so qh is when there's heat input okay so that's usually during the rise of temperature in the system and so when the heat is input okay so and then we also take note that the ideal gas, okay.

01:21 So the we have pv equals to nrt.

01:25 So the product of pv tells us the temperature of the gas.

01:43 So if you look at a to b, a to b is a drop in the product of pv, so there's a drop in temperature.

01:50 B2c is increasing temperature.

01:52 So you have qh.

01:54 C to d is also an increase.

01:55 In temperature because of the increase in product of pv so you have also qh and d to a is a drop in temperature because the product of pv decreases so b to c and c to d is the time where you have qh okay so the qh of the process is equal to q from b to c plus the q for c to d okay so this is equal to b to c is an isochoric process so you have n ccc minus tb and then for qc for process c to d is an isobaric process so the q is ncp t d minus tc okay so i utilize these two formula i so correct process you have q equals to n cv delta t isobaric okay at constant time constant pressure so you use a molahic capacity at constant pressure.

03:01 And then you have our, because this is monotomic, okay, cv equals to 3 over 2r, cp equals to 5 over 2r.

03:13 So you have 3 over 2 and r tc minus tb plus 5 over 2 and r tb minus tc.

03:24 So and then using i do guess equation, okay, and r, rtc is 3 p .0 times we not is the product of pv using i do guess equation.

03:39 Then nrtb is p0, e0.

03:43 Okay.

03:44 And then you do the same thing for the nrtd and rtd will be 6p0, r2, v0 and rtc will be 3p0, we not.

03:55 Okay, so in the end we get the answer in terms of p0 not, which is very nice.

04:00 Okay.

04:00 We have 3 p .0 e0 plus 7 .5 p .0 and we get 10 .5 p .0.

04:09 Okay.

04:12 This is our qh.

04:14 Then our w.

04:17 Okay.

04:17 The w is the area under tb diagram or is actually the area of the rectangle.

04:30 Area of rectangle.

04:34 Okay.

04:36 So if you look at the area of rectangle, you can calculate that...

Please give Ace some feedback