WORK DONE IN VARIOUS THERMODYNAMIC PROCESSES

There are only two ways to transfer energy between a system and its surroundings. 1) heat and 2) work. To increase the energy of a system, heat will be given to the system, and work will be done on the system, and vice versa. Work done is a very important parameter that is considered in any thermodynamic process. There are various thermodynamic processes like isothermal process, adiabatic process, isochoric process, and isobaric process. The possible values of work in each process can be written as.  

Isothermal process. It is a process in which temperature remains constant, that is ΔT =0, the work done in this process, or the value of the amount of energy which can be converted into work, can be calculated as, w= PΔV or dw =PdV (this is an infinitesimal work obtained in any process). The total amount of work obtained in any system can be obtained as, dw = ∫PdV or dw =∫nRT/VxdV (from the general gas equation PV = nRT). As it is an isothermal process, the T remains constant. Thus, integrating only, the variables between two limits, dw =  = dw = nRTln (V₂-V₁) or nRTln (V₂/V₁). To convert ln into common log, this value is obtained as, dw = 2.303nRTlogV₂/V₁. If the value of pressure is given instead of volume, then the value of work is obtained as, dw = 2.303nRTlogP₁/P₂ (as the pressure and volume are inversely proportional to each other).

In a real situation, two types of isothermal processes can occur,

1) Isothermal expansion: In this process, work will be done by the system, so we obtained a negative value of w.

2) Isothermal compression: In this process, work will be done on the system, so we obtain a positive value of w.

Isochoric processes: In this process, the volume change is 0, i.e., ΔV =0. The work done w = PΔV → w = Px0 = 0. Thus, it is concluded that in the case of pressure-volume work, work done can take place only when is volume change occurs. 

Isobaric processes: In these processes, the change in pressure remains constant; the value of work done is obtained as,dw =PdV. The total work obtained is dw =∫PdV → dw =P → dw = P(V₂-V₁) (as the integral of dV is V).

Adiabatic processes: In an adiabatic process, no heat will be exchanged between the system and its surroundings, so dq = 0. In an adiabatic process, the small work done is equal to PdV i.e., w =PdV. The total work done is w =∫PdV. (Here, the relation between P and V cannot be determined by using the general gas equation because P, V, and T are three variables). According to the first law of thermodynamics. dq = dE+dw. As the process is adiabatic, dq =0, thus, dw = -dE → dw = -nCvdT. Total work done dw =    = -nCv (T₂-T₁) → (Cv =R/γ^-1). → dw = -nR/γ^-1(T₂-T₁). → dw = nRT₁/γ^-1 – nRT₂/γ^-1 (As we know that P₁V₁ = nRT₁ and P₂V₂ =nRT₂). → dw = (P₁V₁)^γ = (P₂V₂)^γ →(PV)^γ = constant.

Cyclic process. In a cyclic process, work done cannot be calculated by the expression, i.e.,dw =PdV. Because this expression is used for that process which takes place in one direction only. In a cyclic process, work done can be calculated by using the first law of thermodynamics. dq = dE+dw. As the value of dE is 0, because dE is a state function, and the values of all state functions in a cyclic process are 0. So, dq = dw. Thus, in a cyclic process, total work is the total heat given to the system.

Work done in different processes

There are many types of work.

1. Mechanical work (F x D)
2. Electrical work (Q x ΔV)
3. Gravitational work = m x ΔV
4. Spring work = ½ K(Δx)²
5. Pressure -volume work = PΔV

CARNOT HEAT ENGINE

A Carnot heat engine is a theoretical ideal engine that follows the Carnot laws and operates on the Carnot cycle. It was proposed first time by a French engineer Sadi Carnot, in 1824, and represents the most efficient heat engine possible between two given temperatures (i.e., high temperature and low temperature). It is a reversible cycle, i.e., it can be reversed without any change in the surroundings. It represents the maximum possible efficiency that a heat engine operating between two heat reservoirs (hot and cold) can achieve. No real engine can exceed this efficiency.

Carnot theorem

• No heat engine can be more efficient than the Carnot heat engine operating between the same two temperatures (temperatures of Carnot heat engine and real engine are similar).
• All Carnot heat engines working at the same two temperatures (upper and lower) will have the same efficiency, irrespective of the nature of the substance.
•  The efficiency of the Carnot heat engine depends upon only the temperature of the source and sink. The greater the difference greater the efficiency will be.

Reversible processes in the Carnot cycle

The Carnot engine consists of four stages

1. Isothermal reversible expansion: The process that occurs slowly and by a perfectly conducting base is isothermal expansion. In this process, the working gas or substance expands at constant temperature T₂. To keep the temperature constant, this gas will absorb the heat from the heat source.
2. Adiabatic reversible expansion: In this step, very fast expansion takes place in which the internal energy of the gas decreases. This step is completed by placing the system on a stand (perfectly insulated). The working substance expands without heat transfer, resulting in a decrease in temperature from T₂ to T₁.
3. Isothermal reversible compression: This process is very slow and completed by placing the system in a sink with a perfectly conducting base. The working substance is compressed at a constant temperature of T₁. The extra heat produced in the system is released to the sink.
4. Adiabatic reversible compression: This is a fast process and completed by placing the system on a stand. The working substance is compressed without heat transfer, resulting in an increase in temperature from T₁ to T₂.

According to the 2^nd law of thermodynamics, when any engine works in a cycle (for any engine, it must operate in a cycle) some heat will be lost and some negative work will also be done, which we cannot control.

Work done in different steps of the Carnot heat engine

Isothermal reversible expansion: let T₂, P_1, and V₁ be the temperature, pressure, and volume of a gas at the initial stage. This gas is now allowed to expand isothermally so that volume changes from V₁ to V₂, and pressure changes to P₂. As the process is isothermal and reversible, thus ΔE =0.  In an isothermal process to maintain the temperature, the system must absorb the specific amount of heat q_2, which is utilized for doing work by the system. Thus, q₂ = -w₁ = -nRT₂lnV₂/V₁.

Adiabatic reversible expansion: In this step, the gas is atT₂, P_2, and V₂ conditions of the initial stage. The gas is allowed to expand adiabatically. As q absorbed = 0. Thus, work is done by the gas, and the new temperature is T₁, the pressure is P₃, and the volume is V₃. The work done by gas can be written as: -w₂ = ΔE = – C_V(T₂-T₁).

Isothermal reversible compression: In this step, the gas is compressed reversibly at temperature T₁ from volume V₃ to V₄. During this compression, heat will be produced as q_1, which is transferred to the sink, and the gas remains at constant temperature. As the process is isothermal, therefore ΔE =0.  And the amount of work done can be calculated by using the q and w relations. -q = w₃ = -RT₁ lnV₄/V₃.

Adiabatic reversible compression: Now, in this step, the gas with volume V₄ and temperature T₁ is compressed adiabatically and reversibly and is brought to its initial state. At this stage, the volume of the system becomes V₁ and its temperature T₂. The work done on the system w₄ is given as, w₄ = Cv (T₂-T₁)

Total work done in one cycle

The total work done (w) in one complete cycle is equal to the sum of all work done in all four operations.

W = -w₁ + (-w₂) + w₃ + w₄

= -RT₂ lnV₂/V₁ – Cv (T₂-T₁) – RT₁ ln V₄/V₃ + Cv (T₂-T₁)

W = -RT₂ lnV₂/V₁ + RT₁lnV₃/V₄ = q

Net heat absorbed in one cycle

If q is the net heat absorbed in the whole cycle, then

q = q₂-q₁, where q₂ is the heat absorbed by the system in the isothermal expansion process and q₁ is the heat released in the isothermal compression process. Thus,

q = -RT₂ lnV₂/V₁ – RT₁lnV₄/V₃ or RT₂ lnV₂/V₁ + RT₁lnV₃/V₄

According to the expression used in the adiabatic changes

T₂/T₁ = (V₃/V₂)^γ-1 and T₂/T₁ = (V₄/V₁)^γ-1.

Therefore, V₃/V₂ = V₄/V₁ or V₂/V₁ = V₃/V₄

Substituting the values of V₃/V₄ we get

q = -RT₂ lnV₂/V₁ + RT₁lnV₂/V₁ → -R (T₂-T₁) ln V₂/V₁

As in a cyclic process, the total work done is equal to the net heat absorbed.

therefore, w =  -R (T₂-T₁) ln V₂/V₁     

Efficiency of an engine

The efficiency of an engine is defined as the ratio of the work done by the engine in a cyclic process to the heat absorbed q₂ from the high temperature.

Thus, e = w/q₂ → (-R (T₂-T₁) ln V₂/V₁)/-RT₂ lnV₂/V₁ → e = T₂-T₁/T₂  → 1-T₁/T₂.