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### Course: Thermal physics (Essentials) - Class 11th>Unit 3

Lesson 7: Efficiency of a Carnot engine

# Carnot cycle and Carnot engine

Introduction to the Carnot cycle and Carnot heat engine. Created by Sal Khan.

## Want to join the conversation?

• What's the main purpose of the Carnot Cycle though??
• Sadi Carnot introduced the Carnot cycle in an analysis of the efficiency of heat engines in the early 19th century. He showed that efficiency was lost whenever heat engines deviated from being in thermal equilibrium and that any heat engine operating between a maximum temperature, T1, and a minimum temperature, T2, could not have greater efficiency than a Carnot cycle operating between the same temperatures. So in this sense the Carnot cycle is the theoretical ideal.
I know that PV = nRT.
I'm guessing that the temperature is a function of pressure AND volume T(P,V), but what is the relationship?
• P(V)^n=Constant where n is called the adiabatic exponent IF the process is ADIABATIC For an adiabatic process
n=cp/cv. where cp is the specific heat at constant pressure and cv is the specific heat at constant volume....... Notice that when n=0 ,u get a const pressure line (horizontal). ,when n =1 u get a isothermal curve (rectangular hyperbola) and when n= infinity u get a constant volume curve (vertical line on PV graph.)... Now for adiabatic process ,lets say for system of air, the adiabatic exponent equals 1.40 whixh is >1 and hence the adiabatic curve is MORE sloping down than isothermal curve for air.
• From B to C , how come its follows a ln curve even though temperature isn't constant ?

• The curve from A to B is an isotherm, meaning it's temperature is contstant.
The curve from B to C is adiabatic, which is a different curve. (you can see that is doesn't follow the isotherm.)
That it is a curve is because the system can not jump from one volume to another.

Hope this helped.
• for the D to A process, it's adiabatic and u add pebbles to reach state A, but how do u know it will reach state A exactly instead of somewhere else. i mean how do you make sure it will be the same volume after u add the pebbles
• One has to choose the state, D, very carefully. In other words, the isothermal compression from state C must proceed only until the system reaches the intersection of the isothermal curve for temperature, T2, and the adiabatic curve coming down from state A. That intersection defines the state, D.
• Is this how a reverse cycle air-conditioner works? With the inside being a reservoir and the out side being the other reservoir so for cooling the outside half is compressed allowing the heat to be drawn out side and for heating the inside is compressed allowing the heat to flow into the room?
• Hello Michael,

Correct:

Summer: the evaporator is inside the house

Winter: the evaporator is outside the house

Most systems will use the same coils for summer and winter but will include valves (solenoids) to reverse the connections to the compressor.

At home I have on of these heat pumps. On a cold winter day the outside coil is seriously cold. You would not want to stand in front of it!

Regards,

APD
• At when we start adding pebbles to the system to move from state D to state A was it an adiabatic process?
• The short answer is yes.
B to C and D to A are both adiabatic processes.
A to B and C to D are both isothermic processes.
• Although I intuitively feel that Q1-Q2=W,I still doubt this looking at the PV chart. The area under curve AB =Q1, similarly area under curve CD=Q2 but it is not evident from graph that Q1-Q2=W since curve CD is slightly shifted towards right than AD. How then can we substract? Is there a proof that the substraction yeilds work done?
• Look at stages 2 (going from B to C) and 4 (D->A) of the cycle. I claim that the work done by the system in stage 2 is the same as the work done to the system in stage 4.

Since these processes are adiabatic we have Q=0 and thus:

delta_U = W

Now note that A and B lie on the same isotherm so they have same internal energy. Recall the video where Sal proved that U=3/2nrT. Let's call this internal energy U_1. By the same argument the internal energy at C and D are the same. Let's call this U2.

So we have

U_2 - U_1 = W (stage 2)
U_1 - U_2 = W (stage 4)

proving that the total work for these two stages combined is indeed 0.
• Doesn't an adiabatic process violate the 1st law of thermodynamics ?? If the system is isolated and cant exchange heat with its surroundings and if the KE is used up in doing work , how does the system regain that energy ?? Is it converted to Potential Energy ?
• Adiabatic means that the heat exchange, Q, is 0, not that the internal energy change is 0 (internal energy change is 0 for the isothermal processes). If the process is adiabatic, Q=0, so you can work out the change in internal energy (and therefore temperature) using the 1st law:

dU=dQ-dW
dU=-dW

So the energy for the work done comes from the KE of the particles being used up like you say, and the gas just has a lower internal energy (=lower temperature) at the end of the process.
• Why is pv work done by the system -ve in chemistry and +ve in physics?
• That's more of a semantics difference than an actual distinction in definition. You can say ∫PdV is +(work done by the system) or -(work done on the system). It doesn't depend on whether you're studying physics or chemistry, as well. I've had classes with teachers of different disciplines using the same convention, its something kind of personal haha. My tip: use the one you feel more confortable with and stick to it, while keeping in mind what I wrote on the second sentence "You can say ∫PdV is the +(work done by the system) or -(work done on the system)".
• Can space be considered a reservoir? if so how come the earth and all the other planets didn't match its temperature?
• One aspect of a reservoir is being able to transfer heat to and from it, space being almost empty of matter has no significant ability to receive or provide heat.