A system consisting of n mol of an ideal gas undergoes two...

A system consisting of $n$ mol of an ideal gas undergoes two reversible processes. It starts with pressure $P_{i}$ and volume $V_{i},$ expands isothermally, and then contracts adiabatically to reach a final state with pressure $P_{i}$ and volume 3$V_{i}$. (a) Find its change in entropy in the isothermal process. The entropy does not change in the adiabatic process. (b) What If ? Explain why the answer to part (a) must be the same as the answer to Problem 66.

Key Concepts

Reversible Processes

Reversible processes are idealized thermodynamic processes carried out infinitely slowly so that the system is always in equilibrium. This concept is important because it allows the use of precise mathematical relationships such as dS = dQ_rev/T. In a reversible process, no additional entropy is created, making these processes essential for theoretical derivations of changes in state functions like entropy.

Isothermal Processes

An isothermal process is one during which the temperature of the system remains constant. For ideal gases, this is especially significant because the internal energy does not change, and the heat absorbed or released by the system is exactly balanced by the work done during the expansion or compression. This process is often used to compute entropy changes by integrating dQ_rev/T.

Adiabatic Processes

Adiabatic processes are thermodynamic processes where no heat is exchanged between the system and its surroundings. For reversible adiabatic processes, this means that not only is the heat transfer zero but also the change in entropy is zero. This concept is crucial when analyzing composite processes, such as combining isothermal and adiabatic processes in thermodynamic cycles.

Entropy

Entropy is a fundamental state function in thermodynamics that quantifies the degree of disorder or randomness in a system. Changes in entropy can be calculated using the reversible form of the second law of thermodynamics, where the integrated heat transfer over temperature (dQ_rev/T) determines the net change. Its property as a state function implies that its change depends only on the initial and final states and not on the process path taken.

State Functions and Process Independence

State functions, such as entropy, depend solely on the current state of a system rather than the path taken to reach that state. This concept ensures that when a system undergoes multiple processes (like isothermal and adiabatic transitions) to return to a particular state, the net change in a state function (entropy, in this case) is determined only by the initial and final states. This is why the result for the entropy change in the isothermal expansion must be consistent with other problem solutions involving the same state change.

Transcript

00:01 This question, there's a system with nmo of an ideal gas undergoes two reversible processes.

00:07 The first is isothermal expansion, pivi from piv to pfvf.

00:14 Second is adabetic compression from pfvf to p .i.

00:20 3vi.

00:21 Okay, so from on the pv diagram it will look something like this.

00:29 Okay, isothermal expansion and then adibetic compression.

00:39 Okay, so this is pi, vi, and then here is 3vi, and then here is vf, here is pf.

00:57 So in part a, we need to find a change in entropy in the isothermal process.

01:02 Okay, so before we do anything, okay, so we need to, okay, so the delta s of the isothermal process, it is equal to q over t, okay, because constant temperature, so there's the t is constant, so you need to find q.

01:38 So to find q, to find q, okay, be using first law of thermodynamics.

01:59 Okay, so delta u, delta e internal, it goes to q minus w, q plus w, okay? yeah, i can do both plus or minus.

02:13 Okay, so this w is work done on gas, okay? and then isothermal process means that delta e internal is zero okay so q equals to minus w okay so we need to find w w equals to minus pdv this is equals to minus nrt natural log v final divide by v initial now the task becomes what is vf over vi.

03:04 And then, yeah.

03:11 So you also know that isothermal process, p -i -v -i equals to p -f -v -f, okay, we have a diabetic process.

03:32 We have p -f -v -f to the gamma is equal to pi, tree vi to the gamma.

03:44 Okay, so looking at this, we can do p -i -v -i -v -f to the gamma minus 1...

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