Work, heat, thermodynamics Cylindric Piston

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The discussion revolves around calculating work done by a gas in a cylindrical piston under varying conditions. The initial calculations for work done were found to be incorrect due to the assumption of constant pressure, which is not applicable in this scenario. Participants emphasized the need to consider the type of compression (isothermal, adiabatic) and the path taken during the process to accurately determine work using the integral of pressure with respect to volume. There was confusion regarding the problem's intent, as it lacked clarity on whether the pressure change was gradual or sudden. The need for additional information from the textbook was highlighted to resolve the discrepancies in the calculations.
alingy1
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Consider 5.5 L of a gas at a pressure of 3.0 atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 L.
a) Calculate the work done, and indicate the correct sign.
b) Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 L. The second step results in a final volume of 10.5 L. Calculate the work done, and indicate the correct sign
c) Calculate the work done if after the first step the volume is 8.0 L and the second step leads to a volume of 10.5 L.

a)w=-7.86J
b)w=-9J
c)w=-9.085J

Could you please check my answers? I have a test on thermodynamics soon.

Basically, I used w=-pdeltaV.
For example,
a)p=5.5L*3.0atm/10.5L
deltaV=10.5L-5.5L
w=-pdeltaV=-7.86J.

However, I am not sure for b) and c). The textbook seems to convey that I should get different answers for b) and c). But, they are very similar!
 
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alingy1 said:
Basically, I used w=-pdeltaV.
That equation is only valid for constant P, which is clearly not the case here.
 
Then how do i solve it?
 
alingy1 said:
Then how do i solve it?
You have to use
$$
w = - \int_{V_i}^{V_f} P \; \mathrm{d} V
$$
where ##P## will be a function of ##V##. For an ideal gas, you substitute ##P## using the ideal gas law (if the temperature is constant and can be taken out of the integral).

I notice that no details are given about the kind of compression being done (isothermal, adiabatic, ...). If it's adiabatic compression, you have to use the first law to get ##\Delta U## from which you can deduce ##W##.

Also, I don't understand why doing the compression in steps can change the total work done.
 
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This question comes from Chemistry, Zumdahl, edition 9 in the chapter 6, problem 6/7.(for those who may have the book)

We never saw the types of compression and cal II is not prereq for my chemistry course.

Here is an example of a similar problem we did in class:
Find work done by gas in cylindrical piston:

Vi=0.015L
Vf=0.025L
Pexternal=1atm
W=-1atm*(0.025L-0.015L)

Does this make sense?
 
alingy1 said:
This question comes from Chemistry, Zumdahl, edition 9 in the chapter 6, problem 6/7.(for those who may have the book)
I don't have that book. If it's not too much work to post the full problem, it would help figure out what is asked, as there seems to be information missing.

alingy1 said:
Here is an example of a similar problem we did in class:
Find work done by gas in cylindrical piston:

Vi=0.015L
Vf=0.025L
Pexternal=1atm
W=-1atm*(0.025L-0.015L)

Does this make sense?
Yes. You see, here the external pressure is constant, therefore you can treat it as a constant (!) and look only at the change in volume.
 
This is the whole question! It seems quite disappointing to be honest with you. It is definitely not clear.
 
alingy1 said:
This question comes from Chemistry, Zumdahl, edition 9 in the chapter 6, problem 6/7.(for those who may have the book)

DrClaude said:
I don't have that book. If it's not too much work to post the full problem, it would help figure out what is asked, as there seems to be information missing.

I have this book -- though mine is 6th ed., but it has the same problem. I confirm that the problem statement was posted completely and accurately as given in the book. And I agree that we need information about the type of path taken to solve the problem. Isothermal? A sudden drop to the final pressure, followed by expansion to the final volume? Something else? We simply don't know.

I will note that the question comes from a section called "In-Class Discussion Questions", which according to the book "are designed to be considered by groups of students in class." This is in contrast to questions and exercises that would typically be given as homework.

Since integral calculus is not required for the class, I presume the work is intended to be calculated using simple geometric area formulas to get "the area under the curve" in a P-V diagram. But, no information is given on what shape to use for the curve.
 
your answers are wrong for a and b. those units should be L x atm, if you wanted the answers in joules then you would need to multiply your answers by 101.3
 
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The answer to this problem depends on the original intent of the person who made it up. If the intent was for the external pressure to be changed gradually (reversibly), then the approach recommended by @DrClaude is appropriate. If the intent was for the external pressure to be dropped suddenly to a lower value and to then allow the gas to re-equilibrate at this new pressure, then the approach being used by @alingy1 is appropriate. Since subsequent parts of the problem involve two steps, it seems to me that the latter was the original intent. But, who really knows what the person who made it up was thinking.
 
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