# Is this reaction reversible?

## Homework Statement

Hello, new here! I have a question regarding reversible or irreversible expansions in regards to figuring which equation is needed to figure out work, W. For starters, here is the question:
One mole of an ideal monatomic gas is expanded from an initial state of 3 atm and 500 K to a final state of 1 atm and 300 K. Calculate w, q and ∆U.
Note: Approximate 1 atm  1 x 105 Pa. For an ideal monatomic gas, Cv = 3R/2 and Cp = CV + nR.

So below, I know of two different equations for w,

## Homework Equations

dw = − pex × dV for an irreversible expansion and

Wrev =−nRT Vf ∫ Vi 1/v dV=−nRTln Vf/Vi for a reversible expansion.

## The Attempt at a Solution

I have looked at a few factors in the question, most research points to w = -Pex X dv but no explanation as to why. I also know it is not isothermal or isobaric based on the info above.[/B]
Since the question did not specify reversible or irreversible, which should I use and how should I know how to understand for future reference? Thanks for the help!

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stevendaryl
Staff Emeritus
If you are only given the initial state and final state, then you can't compute the work, because the work depends on the path (that is, it depends on how you get to the final state). So the question must assume that there is some standard way to do the expansion.

Typically, there are three types of expansion that are usually discussed (although you could certainly have weirder types):
1. Isentropic: You expand the gas reversibly, not allowing heat to enter or leave.
2. Isothermal: You expand while keeping the temperature constant.
3. Isobaric: You expand while keeping the external pressure constant.
Neither of your formulas correspond to #1, and #2 is out, because the temperature changes. So I think that the expansion must be #3, isobaric. You seem to think that it can't be isobaric, since the pressure changes from 3 atm to 1 atm, but irreversible isobaric means that the external pressure is kept constant, not the internal pressure. Think of having a container of gas under 3 atm pressure, then you open the container into the room. The external pressure is 1 atm the whole time, while the internal pressure decreases from 3 to 1 atm.

So if it's isobaric, then the work done is $P_{ext} \int dV = P_{ext} (V_{final} - V_{initial})$. You can calculate $V_{final}$ and $V_{initial}$ from $P_{final} V_{final} = N R T_{final}$ and $P_{initial} V_{initial} = N R T_{initial}$

Chestermiller
Mentor
If you are only given the initial state and final state, then you can't compute the work, because the work depends on the path (that is, it depends on how you get to the final state). So the question must assume that there is some standard way to do the expansion.

Typically, there are three types of expansion that are usually discussed (although you could certainly have weirder types):
1. Isentropic: You expand the gas reversibly, not allowing heat to enter or leave.
2. Isothermal: You expand while keeping the temperature constant.
3. Isobaric: You expand while keeping the external pressure constant.
Neither of your formulas correspond to #1, and #2 is out, because the temperature changes. So I think that the expansion must be #3, isobaric. You seem to think that it can't be isobaric, since the pressure changes from 3 atm to 1 atm, but irreversible isobaric means that the external pressure is kept constant, not the internal pressure. Think of having a container of gas under 3 atm pressure, then you open the container into the room. The external pressure is 1 atm the whole time, while the internal pressure decreases from 3 to 1 atm.

So if it's isobaric, then the work done is $P_{ext} \int dV = P_{ext} (V_{final} - V_{initial})$. You can calculate $V_{final}$ and $V_{initial}$ from $P_{final} V_{final} = N R T_{final}$ and $P_{initial} V_{initial} = N R T_{initial}$
This is certainly a viable path, consistent with the problem statement. Another interesting possibility would be to assume a polytropic (reversible) path described by $pV^n=Const$, where n can be determined from the initial and final pressures and volumes. It would be interesting to see how these two results would compare.