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

In summary: I have calculated the work done by the gas on the surroundings using the method you introduced, and it matches the answer from the textbook. I wasn't sure of the log as Vf<Vi, but the log is negative (as expected given what was mentioned above).In summary, a cylindrical container with 0.200 mol of oxygen, sealed by a movable piston, undergoes a series of transformations. The initial pressure is 2.5 × 10^5 Pa and the initial temperature is 77◦C, with an ideal gas constant of R = 8.315J/mol · K. First, the oxygen undergoes an isobaric expansion to twice its original volume, then it is compressed isothermally back
  • #1
BOAS
552
19

Homework Statement



A cylindrical container is sealed with a movable piston and contains [itex]0.200[/itex] mol of oxygen. The initial pressure is [itex]2.5 × 10^5[/itex] Pa and the initial temperature is [itex]77◦C[/itex]. The value of the ideal gas constant is [itex]R = 8.315[/itex]J/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 [itex]W = P(\Delta V)[/itex], and the work done by the isothermal process is given by [itex]W = nRT ln(\frac{V_{f}}{V_{i}})[/itex].

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 [itex]W = nRT ln(\frac{V_{f}}{V_{i}}) - P(\Delta V)[/itex]?

Thanks!
 
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  • #2
BOAS said:
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 [itex]W = nRT ln(\frac{V_{f}}{V_{i}}) - P(\Delta V)[/itex]?
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?
 
  • #3
vela said:
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?
 
  • #4
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.
 
  • #5
There is also an issue with the subscripts on the volumes. If Vi is the initial pressure before the isobaric expansion, and Vf is the gas volume after the expansion, then during the isothermal compression, the starting volume is Vf and the ending volume is Vi. 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
 
  • #6
thanks for the help, I think I have taken it all into account.
 

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

1. What is work done in a PV diagram?

Work done in a PV diagram is the area under the curve on the graph. It represents the amount of energy transferred to or from a system due to a change in volume at a constant pressure.

2. How is work calculated in a PV diagram?

Work is calculated by multiplying the magnitude of the force acting on an object by the distance the object moves. In a PV diagram, this can be represented by finding the area under the curve on the graph.

3. What does a positive work value indicate in a PV diagram?

A positive work value in a PV diagram indicates that work is being done on the system, meaning energy is being transferred into the system. This can occur when the volume of a system is increasing at a constant pressure.

4. Can work done be negative in a PV diagram?

Yes, work done can be negative in a PV diagram. This indicates that work is being done by the system, meaning energy is being transferred out of the system. This can occur when the volume of a system is decreasing at a constant pressure.

5. How can work done be used to determine the efficiency of a system?

Work done can be used to determine the efficiency of a system by comparing the amount of work input to the amount of work output. The efficiency of a system is the ratio of work output to work input, and a higher efficiency indicates that less energy is lost during the process.

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