Hey,
Can someone clarify what to do first in this problem? Thanks!
Calculate the work for each of the following processes beginning with a gas sample in a piston assembly with T = 305 K , P = 1.79 atm, and V = 4.29 L: (a) irreversible expansion against a constant external pressure of 1.00 atm to a final volume of 6.52 L; (b) isothermal, reversible expansion to a final volume of 6.52 L.
Textbook 4B.13
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 152
- Joined: Fri Sep 24, 2021 6:19 am
Re: Textbook 4B.13
For this kind of problem, approaching it from the reversible isothermal side is easier to start with for me. We know that reversible isothermal expansion always does the maximal amount of work, so it's easy after that to compare your answer and see if it's right.
The equation for rev. iso. expansion is w=-nRT*ln(Vf/Vi) (so if it's expanding, work is negative, and the system is doing work on the surroundings).
Therefore, by plugging in n (find the # of moles by using PV=nRT for the initial condition), R, T, and the initial and final volumes, you can find the work done pretty directly.
The other process (not rev. iso.) can be more complex. But in this case, for irreversible expansion against constant pressure, we just use the equation w=-Pext*(deltaV) = -Pext*(Vf-Vi), which you can also get from the equation. This value for w should be SMALLER in magnitude than the work done by rev. iso. expansion.
Overall, the difference in the two shows that work is not a state function, and it depends on the path taken.
The equation for rev. iso. expansion is w=-nRT*ln(Vf/Vi) (so if it's expanding, work is negative, and the system is doing work on the surroundings).
Therefore, by plugging in n (find the # of moles by using PV=nRT for the initial condition), R, T, and the initial and final volumes, you can find the work done pretty directly.
The other process (not rev. iso.) can be more complex. But in this case, for irreversible expansion against constant pressure, we just use the equation w=-Pext*(deltaV) = -Pext*(Vf-Vi), which you can also get from the equation. This value for w should be SMALLER in magnitude than the work done by rev. iso. expansion.
Overall, the difference in the two shows that work is not a state function, and it depends on the path taken.
-
- Posts: 102
- Joined: Fri Sep 24, 2021 6:52 am
Re: Textbook 4B.13
Hi, this question involves knowing what's happening in the problem and what equations can be used. For a, since it is irreversible expansion against a constant external pressure, we can use the equation w = -Pex x change in volume. Plugging in the values, we would get w = -(1.00 atm) x (6.52-4.59 L), which equals -2.23 L atm, which we would then multiply by our conversion factor of 101.325J/Latm to get the work in Joules. For b, we would use the equation w = -nRTln(V2/V1). You would first have to find the number of moles of the gas using PV=nrT. From there, you would follow a similar process of plugging in values to find the work.
Re: Textbook 4B.13
Would the reversible portion of the question be asked on the test because on outline 3 it only says:
"Calculate the work of expansion at constant pressure, w = - V1∫V2 PEX dV = - PEX ∆V"
This does not seem to capture the idea we would have to know how to calculate the work of expansion at changing pressure.
"Calculate the work of expansion at constant pressure, w = - V1∫V2 PEX dV = - PEX ∆V"
This does not seem to capture the idea we would have to know how to calculate the work of expansion at changing pressure.
-
- Posts: 52
- Joined: Wed Nov 18, 2020 12:27 am
Re: Textbook 4B.13
Hey Eddie,
You're definitely right that the outline only mentions "constant pressure", but I'd still study irreversible expansion just in case. If it's on the homework, it's probably fair game for Lavelle's tests.
You're definitely right that the outline only mentions "constant pressure", but I'd still study irreversible expansion just in case. If it's on the homework, it's probably fair game for Lavelle's tests.
Return to “Phase Changes & Related Calculations”
Who is online
Users browsing this forum: No registered users and 9 guests