Consider an infinitesimal reversible adiabatic compression...

Consider an infinitesimal reversible adiabatic compression or expansion process. By taking $s=s(P, v)$ and using the Maxwell relations, show that for this process $P v^{k}=$ constant, where $k$ is the isentropic expansion exponent defined as $$k=\frac{V}{P}\left(\frac{\partial P}{\partial V}\right)$$ Also, show that the isentropic expansion exponent $k$ reduces to the specific heat ratio $c_{p} / c_{v}$ for an ideal gas.

Key Concepts

Specific Heat Ratio

The specific heat ratio (often denoted as ? or k) is defined as the ratio of the specific heat at constant pressure to that at constant volume (cp/cv). For ideal gases, this ratio plays a crucial role in characterizing the behavior of the gas in adiabatic processes. It directly appears in expressions that relate the thermodynamic variables during an isentropic process, and its value determines the exponent in the relationship between pressure and volume.

Isentropic Process

An isentropic process is one that occurs at constant entropy. This condition is a direct consequence of a reversible adiabatic process and underpins many important relationships in thermodynamics, such as the derivation of the adiabatic equation PV^k = constant. Because the entropy does not change, the process can be characterized by specific power laws relating pressure and volume that simplify the analysis of work and energy transfer in the system.

Maxwell’s Relations

Maxwell’s relations are a set of equations derived from the equality of mixed partial derivatives of thermodynamic potentials. They connect different measurable thermodynamic quantities and are especially useful in relating entropy derivatives to other state functions like pressure and volume. In the context of an adiabatic process, using Maxwell’s relations allows us to express differential changes in entropy as functions of variables like pressure and volume.

Reversible Process

A reversible process is an idealized thermodynamic process that happens infinitely slowly, ensuring the system remains in thermodynamic equilibrium at all points. This allows the process to be reversed without any net change in both the system and its surroundings. The reversibility assumption simplifies the derivation of thermodynamic relationships by ensuring that the usual equations of state and thermodynamic identities, such as Maxwell's relations, hold.

Adiabatic Process

An adiabatic process is one in which no heat is transferred into or out of the system––all changes in internal energy are solely due to work done by or on the system. In thermodynamics, when a process is both adiabatic and reversible, it is isentropic, which means the entropy remains constant. This concept is crucial for deriving relationships between state variables such as pressure and volume during the process.

Transcript

00:01 In this question we need to show two things.

00:02 First thing is for isentropic process, pv to the k because a constant where k is minus l over p, partial p partial v in fixed s.

00:12 And then for i do guess, k equals to cp over cv.

00:17 Okay, so to show the first part, so we notes that for isentropic process, the s equals to 0.

00:35 And for s as a function of pressure and specific volume, the total differential for specific entropy will be written as partial s, partial p, fixed v, d .p plus partial s, partial v, fixed p, dv.

01:01 And this is equal to 0.

01:04 Okay.

01:05 And then next we'll be using actual relations.

01:14 So partial s, partial p, 6v is equal to minus partial v, partial t, 6s, and then partial s, partial s, partial v and fixed p is equal to partial p, partial t, a fix s.

01:36 Okay, so we'll replace the terms in the equation above.

01:44 So we have minus partial v, partial t, fixed s d p plus partial p partial t, fix s d equal to 0.

01:58 And then the next step, we will do chain rule for the second term.

02:05 So we have partial p, partial v, six s, and then partial v, partial t, six as dv equals zero.

02:17 So here what we are doing is try to, is getting rid of this partial derivative.

02:22 So we have minus dp plus partial p partial v 6 s dv equals to 0...

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