POLYTROPIC PROCESS: CHARACTERISTICS, APPLICATIONS AND EXAMPLES - PHYSICAL - 2025

A polytropic process is a thermodynamic process that occurs when the relationship between pressure P and volume V given by PV ⁿ is kept constant. The exponent n is a real number, generally between zero and infinity, but in some cases it can be negative.

The value of n is called the polytropy index and it is important to note that during a polytropic thermodynamic process, said index must maintain a fixed value, otherwise the process will not be considered polytropic.

Figure 1. Characteristic equation of a polytropic thermodynamic process. Source: F. Zapata.

Characteristics of polytropic processes

Some characteristic cases of polytropic processes are:

- The isothermal process (at constant temperature T), in which the exponent is n = 1.

- An isobaric process (at constant pressure P), in this case n = 0.

- The isochoric process (at constant volume V), for which n = + ∞.

- Adiabatic processes (at constant S entropy), in which the exponent is n = γ, where γ is the adiabatic constant. This constant is the quotient between the heat capacity at constant pressure Cp divided by the heat capacity at constant volume Cv:

γ = Cp / Cv

- Any other thermodynamic process that is not one of the previous cases. but that meets PV ⁿ = ctte with a real and constant polytropic index n will also be a polytropic process.

Figure 2. Different characteristic cases of polytropic thermodynamic processes. Source: Wikimedia Commons.

Applications

One of the main applications of the polytropic equation is to calculate the work done by a closed thermodynamic system, when it passes from an initial state to a final state in a quasi-static way, that is, following a succession of equilibrium states.

Work on polytropic processes for different values ​​of n

For n ≠ 1

The mechanical work W performed by a closed thermodynamic system is calculated by the expression:

W = ∫P.dV

Where P is pressure and V is volume.

As in the case of a polytropic process, the relationship between pressure and volume is:

We have the mechanical work done during a polytropic process, which begins in an initial state 1 and ends in the final state 2. All this appears in the following expression:

C = P ₁ V ₁ ⁿ = P ₂ V ₂ ⁿ

By substituting the value of the constant in the work expression, we obtain:

W = (P ₂ V ₂ - P ₁ V ₁) / (1-n)

In the case that the working substance can be modeled as an ideal gas, we have the following equation of state:

PV = mRT

Where m is the number of moles of the ideal gas and R is the universal gas constant.

For an ideal gas that follows a polytropic process with a polytropy index different from unity and that passes from a state with initial temperature T ₁ to another state with temperature T ₂, the work done is given by the following formula:

W = m R (T ₂ - T ₁) / (1-n)

For n → ∞

According to the formula for the work obtained in the previous section, we have that the work of a polytropic process with n = ∞ is null, because the expression of the work is divided by infinity and therefore the result tends to zero.

Another way to arrive at this result is to start from the relation P ₁ V ₁ ⁿ = P ₂ V ₂ ⁿ, which can be rewritten as follows:

(P ₁ / P ₂) = (V ₂ / V1) ⁿ

Taking the nth root in each member, we obtain:

(V ₂ / V1) = (P ₁ / P ₂) ^{(1 / n)}

In the case that n → ∞, we have (V ₂ / V1) = 1, which means that:

V ₂ = V ₁

That is, the volume does not change in a polytropic process with n → ∞. Therefore, the volume differential dV in the integral of mechanical work is 0. This type of polytropic processes are also known as isochoric processes, or constant volume processes.

For n = 1

Again we have the expression the expression for work:

W = ∫P dV

In the case of a polytropic process with n = 1, the relationship between pressure and volume is:

PV = constant = C

By solving P from the previous expression and substituting, we have the work done to go from initial state 1 to final state 2:

That is to say:

W = C ln (V ₂ / V ₁).

As the initial and final states are well determined, so will the ctte. That is to say:

C = P ₁ V ₁ = P ₂ V ₂

Finally, we have the following useful expressions to find the mechanical work of a closed polytropic system in which n = 1.

W = P ₁ V ₁ ln (V ₂ / V ₁) = P ₂ V ₂ ln (V ₂ / V ₁)

If the working substance consists of m moles of ideal gas, then the ideal gas equation of state can be applied: PV = mRT

In this case, since PV ₁ = ctte, we have that a polytropic process with n = 1 is a process at constant temperature T (isothermal), so that the following expressions for the work can be obtained:

W = m RT ₁ ln (V ₂ / V ₁) = m RT ₂ ln (V ₂ / V ₁)

Figure 3. A melting icicle, example of an isothermal process. Source: Pixabay.

Examples of polytropic processes

- Example 1

Suppose a cylinder with a movable piston filled with one kilogram of air. Initially the air occupies a volume V ₁ = 0.2 m ³ at a pressure P ₁ = 400 kPa. A polytropic process is followed with n = γ = 1.4, whose final state has pressure P ₂ = 100 kPa. Determine the work done by the air on the piston.

Solution

When the polytropy index equals the adiabatic constant, there is a process in which the working substance (air) does not exchange heat with the environment, and therefore the entropy does not change either.

For air, a diatomic ideal gas, we have:

γ = Cp / Cv, with Cp = (7/2) R and Cv = (5/2) R

So:

γ = 7/5 = 1.4

Using the expression of the polytropic process, the final volume of the air can be determined:

V ₂ = ^{(1 / 1.4)} = 0.54 m ³.

Now we have the conditions to apply the formula of work done in a polytropic process for n ≠ 1 obtained above:

W = (P ₂ V ₂ - P1 V1) / (1-n)

Substituting the appropriate values ​​we have:

W = (100 kPa 0.54 m ³ - 400 kPa 0.2 m ³) / (1 - 1.4) = 65.4 kJ

- Example 2

Assume the same cylinder from Example 1, with a movable piston filled with one kilogram of air. Initially the air occupies a volume V1 = 0.2 m ³ at a pressure P1 = 400 kPa. But unlike the previous case, the air expands isothermally to reach a final pressure P2 = 100 kPa. Determine the work done by the air on the piston.

Solution

As seen previously, isothermal processes are polytropic processes with index n = 1, so it is true that:

P1 V1 = P2 V2

In this way, the final volume can be easily detached to obtain:

V2 = 0.8 m ³

Then, using the work expression obtained previously for the case n = 1, we have that the work done by the air on the piston in this process is:

W = P1 V1 ln (V2 / V1) = 400000 Pa × 0.2 m ³ ln (0.8 / 0.2) = 110.9 kJ.

References

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