Thermodynamic Application Isothermal Work

AI Thread Summary
The discussion revolves around solving a thermodynamics problem involving an ideal gas expanding isothermally. The key equations include the work done (W = nRTln(vi/vf)) and the ideal gas law (PV = nRT). Participants express confusion about determining the initial volume and temperature, noting that the initial volume must be less than 25 L. The importance of using correct units for volume is emphasized, as it affects calculations significantly. Ultimately, the participants are encouraged to first calculate temperature using known final pressure and volume before determining the initial volume.
latitude
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Homework Statement



One mole of an ideal gas does 3000 J work on its surroundings as it expands isothermally to a final pressure of 1 atm and volume 25 L. Determine a) initial volume and b) Temp of gas

Homework Equations

The Attempt at a Solution


Well, its isothermal, so Temp is constant, so
W = nRTln(vi/vf)
It says it expands, so the initial volume is <25 L...
that's kinda all I got right now; I'm not sure how to get the next step, since both T and vi are in the equation. It must have something to do with pressure.
 
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latitude said:

Homework Statement



One mole of an ideal gas does 3000 J work on its surroundings as it expands isothermally to a final pressure of 1 atm and volume 25 L. Determine a) initial volume and b) Temp of gas

Homework Equations

The Attempt at a Solution


Well, its isothermal, so Temp is constant, so
W = nRTln(vi/vf)
It says it expands, so the initial volume is <25 L...
that's kinda all I got right now; I'm not sure how to get the next step, since both T and vi are in the equation. It must have something to do with pressure.
How do you find the temperature of the gas? Find that and then determine what ln(vi/vf) has to be if W=3000J.

AM
 
PV = nRT
(1.013 x 10^5)(10 m^3) = (1)(8.314)T
T = Some really big, way illogical number? Unless the moles thing is wrong...

ARGH. This one is frustrating me. I'm wondering if I don't have to find volume first??
 
Last edited:
Oooh...

3000 J = nR(PV/nR)(ln(vi/vf))
3000 J = (10.13 x 10^5)(25 L) ln(vi/25)
1.184 x 10^-3 = vi/25
vi = 0.029 L??

Possible??

EDIT: No, cause my final temp is still 304606.7 K !
 
Last edited:
latitude said:
Oooh...

3000 J = nR(PV/nR)(ln(vi/vf))
3000 J = (10.13 x 10^5)(25 L) ln(vi/25)
1.184 x 10^-3 = vi/25
vi = 0.029 L??

Possible??

EDIT: No, cause my final temp is still 304606.7 K !
Start with PV=nRT

P_iV_i = nRT = P_fV_f

You know P_f, V_f and n so work out T. Be careful to use the correct units for volume. Units are litres or 10^-3 m^3.


Then, from W = nRTln(vi/vf), work out what vi is.

AM
 
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