Finding the work done by a gas under quasi-static, isothermal expansion

OUNT OF GAS REMAINS CONSTANT. You can use n=n in W=nRTln(Pi/Pf).In summary, an ideal gas initially at 293K and 200kPa undergoes a quasi-static, isothermal expansion until its pressure is reduced to 100kPa. The work done by the gas can be determined using the formula W=nRTln(Pi/Pf), where n is the amount of gas which remains constant and can be found using the ideal gas law.
  • #1
jobloggs
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0
1. So, an ideal gas is initially at 293K and 200kPa, and has a volume of .004m3
It undergoes a quasi-static, isotheral expansion until its pressure is reduced to 100kPa
What is the Work done by the gas



2.I know that for isothermal conditions W=nRTln(Pf/Pi)



3. my first attempt was making n=1 and Pf=200kPa and Pi=100kPa
i got W to be 1685w.

My first problem with this is that I don't actually know what n is, and I'm not even sure what Pi and Pf are, can't find it in my lecture notes anywhere.

My next step was to isntead of guessing n=1, make PV=nRt because I have P and V, but I'm not sure which P to use.


I know this dosen't seem like a difficult problem, but i won't be getting the answers until next month, so any guidance will be appreciated
 
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  • #2
jobloggs said:
[
My next step was to isntead of guessing n=1, make PV=nRt because I have P and V, but I'm not sure which P to use. [/b]

Both P and V change during the expansion. You don't know the final volume, but you do know the initial volume. So, use the initial pressure as well.
 
  • #3
jobloggs said:
1. So, an ideal gas is initially at 293K and 200kPa, and has a volume of .004m3
It undergoes a quasi-static, isotheral expansion until its pressure is reduced to 100kPa
What is the Work done by the gas



2.I know that for isothermal conditions W=nRTln(Pf/Pi)
It should be Pi/Pf.

dW = PdV. If you substitute P = nRT/V (where T is constant since it is isothermal) you get:

[tex]\int_{V_i}^{V_f} PdV = nRT\int_{V_i}^{V_f} dV/V = nRT\ln (\frac{V_f}{V_i})[/tex]

Since, by the ideal gas law, PiVi/Ti = PfVf/Tf, Vf/Vi = Pi/Pf (where Ti and Tf are equal)

3. my first attempt was making n=1 and Pf=200kPa and Pi=100kPa
i got W to be 1685w.
My first problem with this is that I don't actually know what n is, and I'm not even sure what Pi and Pf are, can't find it in my lecture notes anywhere.

My next step was to instead of guessing n=1, make PV=nRt because I have P and V, but I'm not sure which P to use.
You have to find n. Use the ideal gas law:

PV=nRT

You know Pi, Vi and Ti so you can find n. n does not change.

AM
 

1. What is quasi-static expansion?

Quasi-static expansion refers to a process in which a gas expands slowly and gradually, with the external pressure remaining almost constant. This allows the gas to adjust to the changing conditions and undergo a reversible process.

2. What does isothermal mean in relation to gas expansion?

Isothermal expansion means that the temperature of the gas remains constant during the expansion process. This is achieved by adding or removing heat to maintain a constant temperature.

3. How is the work done by a gas calculated under quasi-static, isothermal expansion?

The work done by a gas in this scenario can be calculated using the formula W = -nRTln(Vf/Vi), where n is the number of moles of gas, R is the gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume.

4. What factors affect the work done by a gas under quasi-static, isothermal expansion?

The work done by a gas in this scenario is affected by the number of moles of gas, the gas constant, the temperature, and the change in volume. These factors can be manipulated to increase or decrease the work done.

5. Why is it important to consider quasi-static, isothermal expansion in scientific experiments?

Quasi-static, isothermal expansion allows for a more controlled and reversible process, making it easier to measure and accurately calculate the work done by a gas. It also ensures that the gas does not undergo any sudden changes in temperature or pressure which could affect the results of the experiment.

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