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

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SUMMARY

The discussion centers on calculating the work done by an ideal gas during a quasi-static, isothermal expansion from an initial pressure of 200 kPa to a final pressure of 100 kPa at a constant temperature of 293K. The formula for work done under isothermal conditions is W = nRT ln(Pf/Pi). Participants clarified the need to determine the number of moles (n) using the ideal gas law, PV = nRT, where initial values of pressure and volume are known. The correct approach involves substituting the initial conditions to find n before applying the work formula.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of isothermal processes in thermodynamics
  • Familiarity with logarithmic functions for calculations
  • Basic calculus for integrating work done (dW = PdV)
NEXT STEPS
  • Calculate the number of moles (n) using the ideal gas law with given initial conditions
  • Learn about the implications of quasi-static processes in thermodynamics
  • Explore the derivation of work done in isothermal expansions
  • Investigate the relationship between pressure, volume, and temperature in gas laws
USEFUL FOR

Students studying thermodynamics, engineers working with gas systems, and anyone interested in understanding the principles of work done during gas expansions.

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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 doesn'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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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.
 
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:

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

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
 

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