Homework Help: Work, heat, thermodynamics Cylindric Piston

1. Feb 1, 2014

alingy1

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!

2. Feb 1, 2014

Staff: Mentor

That equation is only valid for constant P, which is clearly not the case here.

3. Feb 1, 2014

alingy1

Then how do i solve it?

4. Feb 1, 2014

Staff: Mentor

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.

5. Feb 1, 2014

alingy1

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?

6. Feb 1, 2014

Staff: Mentor

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.

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.

7. Feb 1, 2014

alingy1

This is the whole question! It seems quite disappointing to be honest with you. It is definitely not clear.

8. Feb 2, 2014

Redbelly98

Staff Emeritus
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.

9. Apr 4, 2018

COOO

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

10. Apr 4, 2018

Staff: Mentor

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.