Class 11 Physics Chapter 12 Thermodynamics
Reversible process is one in which system and surrounding can return to their initial states and final states. But reversible process is a theoretical concept. To build such machine. To design such machine having efficiency close to reversible process. Sadi Carnot, a French Engineer tries to find out solution to this problem. In this article we will talk in detail about Carnot cycle.
The Carnot Cycle
Sadi Carnot briefly studied reversible process. He found that, reversible process is no possible. Then he thought if reversible process is not possible can’t we use cyclic process for this purpose. So he designed a cyclic process that goes through different states and come back on the same initial state. He designed a cyclic process that goes through different states and come back on the same initial state. He designed a cyclic process called Carnot Cycle. Carnot cycle is made up of four steps
Heat Q1 supplied from hot reservoir to the gas results in isothermal expansion. Hence gas changes its state, from initial state (P1, V1, T1) to intermediate state (P2, V2, T1). In isothermal process internal energy remains constant and therefore entire heat supplied get converted to work done shown by
Q1 = W1 →2 = μ R T1 ln (V2/V1)
To complete the cycle here we need to change the path of the gas. It can be possible by obstructing the supply of heat and changing the temperature. Thus we need to change the temperature of gas from T1 to T2. For this, we can use adiabatic expansion, in which increase in volume results in fall of temperature to T2. Thus during adiabatic expansion the gas reaches from state (P 2, V 2, T 1) to state (P 3, V 3, T 2)
During this process work done by the gas is
W2→3=(1/1-ƴ) μ RT1-μ RT2
During this process the gas releases heat Q2 to the cold reservoir which is at same temperature i.e. T2. Thus the gas changes its state from (P 3, V 3, T 2) to (P 4, V 4, T 2). During this process work is done on the gas by the surrounding. It is given as
W3→4 = Q2 = μ R T2 ln (V3/V4)
Finally the cycle is completed by adiabatic compression i.e. by doing work on the gas the state of the gas will get changed from (P 4, V 4, T 2) to (P 1, V 1, T 1).The work done on the gas can be calculated as
W4→1=(1/1-ƴ) μ RT1-μ RT2
This cycle is called as Carnot Engine.
Efficiency of carnot engine
The efficiency of carnot engine is given by,
η = 1 – T2/T1
Carnot engine is a cyclic process. It uses two reservoirs hot and cold for completing the cyclic process. Since it is a reversible cyclic process hence each step can be reversed. E.g. if the steps of heat engine are reversed then we will get cyclic process of Refrigerator or vice versa. However from the study of Carnot Cycle we get very important result which is known as Carnot’s Theorem. It states that “All reversible engines operating between the same two temperatures, always have equal efficiency but no engine can have efficiency more than the Carnot engine.”
Let us prove Carnot Theorem now. Consider two engines, a reversible engine denoted by R and an irreversible engine denoted by I. Suppose these are working between the hot and cold Reservoir. We know that reversible engine can act in any direction.
Let’s assume it acts as Refrigerator, and Irreversible engine act as Heat Engine. Here, we have to prove that irreversible engine I, could not have greater efficiency than reversible engine ‘R’. To prove this theory, Let us make an assumption that efficiency of irreversible engine is greater than reversible engine. I.e. ηI > ηR. Suppose the irreversible engine is a heat engine and reversible engine is set to act as refrigerator. The irreversible heat engine I extract heat Q1 from hot reservoir and convert it to work W’ releasing heat Q2’ = (Q1-W’) to cold reservoir. Similarly reversible heat engine extract heat (Q2) from cold reservoir when the work W is done on it and release heat Q1 to hot reservoir. With this respect Q2 = Q1 – W. Since ηI > ηR hence work done by irreversible engine must be greater than work done on reversible engine. I.e. W’>W with this respect can we say that Q2>Q2’ yes definitely. Let us think this all together. The combined working of the arrangement will be such that the system extract heat Q2 – Q2’ from cold reservoir and convert it into equivalent work
Q2 – Q2’ = (Q1 – W) – (Q1 – W’) = W’ – W
This state that the heat absorbed from cold reservoir (Q2 – Q2’) totally get converted to work (W’ – W). But according to Second Law of Thermodynamics stated by Clausius, i.e. We can not extract heat from cold reservoir without doing external work. Hence the arrangement is not possible. It means that our assumption ηI > ηR is wrong. If it is wrong, then the right assumption will be, efficiency of irreversible engine can not be greater than reversible engine.i.e. ηI < ηR. And So, Carnot Theorem is Verified.
There is one more important result from the expression of Efficiency of Carnot engine
η = 1 – T2/T1
This shows that,. the efficiency of the Carnot engine is Independent of nature of the substance. Also, efficiency of Carnot engine can be expressed in terms of ratio of heat absorbed and released during Isothermal Processes as η = 1 – Q2/Q1 . So from the two equations of efficiency of Carnot engine we get a Universal Equation given as
Q1/Q2 = T1/T2
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Keywords: Carnot engine, Carnot theorem, Isotheorem process, Adiabatic process,