Is the Work Done by the Piston on the Gas Counted Twice in PV Diagrams?

[BOAS]

Homework Statement

A cylindrical container is sealed with a movable piston and contains 0.200 mol of oxygen. The initial pressure is 2.5 × 10^5 Pa and the initial temperature is 77◦C. The value of the ideal gas constant is R = 8.315J/mol · K. The oxygen, which can be approximated as an ideal gas, first undergoes an isobaric expansion to twice its original volume. It is then compressed isothermally back to its original volume. Finally, it is cooled isochorically to its original pressure.

(a) Give a definition of isobaric, isothermal, and isochoric transformations. Show the series of processes on a p-V diagram. Label your diagram clearly.
(b) Compute the temperature during the isothermal compression.
(c) Compute the maximum pressure.
(d) Compute the total work done by the piston on the gas during the series of processes.
(e) Compute the oxygen’s internal energy change during the initial isobaric expansion. Use CP = 29.17 J/mol · K. (f) Compute the oxygen’s internal energy change during the isothermal compression.

Homework Equations

The Attempt at a Solution

I have done parts a, b and c without a problem, but I have a small question regarding part d.

The work done by the isobaric process is W = P(\Delta V), and the work done by the isothermal process is given by W = nRT ln(\frac{V_{f}}{V_{i}}).

The work done by the isothermal process seems to count the work done during the Isobaric process twice. Is this correct, or would the work done by the Isothermal process in this case be W = nRT ln(\frac{V_{f}}{V_{i}}) - P(\Delta V)?

Thanks!

 

[vela]

The work done by the isothermal process seems to count the work done during the Isobaric process twice. Is this correct, or would the work done by the Isothermal process in this case be W = nRT ln(\frac{V_{f}}{V_{i}}) - P(\Delta V)?

Be careful with the signs. When the gas is allowed to expand, $p\Delta V$ is a positive number. Is that work done on the gas or on the surroundings? Similarly, is your expression for work done during the isothermal compression work done on the gas or on the surroundings?

 

[BOAS]

Be careful with the signs. When the gas is allowed to expand, $p\Delta V$ is a positive number. Is that work done on the gas or on the surroundings? Similarly, is your expression for work done during the isothermal compression work done on the gas or on the surroundings?

The work being done is by the system on it's surroundings - I can see I've muddled this a few times.

So $p\Delta V$ does work on the piston.

The work done during the isothermal compression should be $W = -nRTln(\frac{v_{f}}{v_{i}})$ since the piston does work on the gas. Here, I'm a little unsure. The piston surely must do work to compress the gas, but the logarithm < 1 so overall it's a positive quantity. Do I place a negative sign by convention, or keep it as a positive value?

 

[vela]

One approach is to pick a convention and be consistent. If you go with the definition
$$W = \int p\,dV$$ then W represents the work done on the surroundings by the gas. For an isobaric process, you then have $W = p\Delta V$, and in the case of an isothermal process, you have
$$W = \int \frac{nRT}{V}\,dV = nRT \ln \frac{V_f}{V_i}.$$ The other convention is to define
$$W = -\int p\,dV$$ where W represents the work done on the gas by the surroundings. For the isobaric process, you have $W = -p\Delta V$. For the isothermal process, you have $W = -nRT \ln\frac{V_f}{V_I}$. In either case, you just add the contributions together. Since the problem asks for the work done on the gas, if you go with the first convention, you have to flip the sign of W to get the answer to the question.

 

[Chestermiller]

There is also an issue with the subscripts on the volumes. If V_i is the initial pressure before the isobaric expansion, and V_f is the gas volume after the expansion, then during the isothermal compression, the starting volume is V_f and the ending volume is V_i. You need to make sure that this is properly taken into account in determining the isothermal work that the gas does on the surroundings during the isothermal expansion.

Chet

 

[BOAS]

thanks for the help, I think I have taken it all into account.