Expansion of a gas at constant temperature and pressure

AI Thread Summary
The discussion focuses on calculating the work done during the expansion of a gas at constant temperature and pressure. For expansion against a vacuum, no work is done as there is no opposing force. When expanding against constant pressures of 1.5 atm and 2.8 atm, the work is calculated using the formula w = -PdV, with necessary conversions from mL to cubic meters. It is noted that for isothermal expansion, the pressure varies with volume, requiring the use of the integral W = -∫ P dv for accurate calculations. Proper unit conversion to the MKS system is emphasized for consistency in calculations.
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Homework Statement



A gas expands in volume from 26.7mL to 89.3mL at constant temperature. Calculate the work done (in joule) if the gas expands:
i) against a vacuum
ii) against constant pressure of 1.5atm
iii) against a constant pressure of 2.8


Homework Equations



w= -PdV

gas expands at constant temperature= isothemic expansion

The Attempt at a Solution



i) I'm not sure about this part

ii) 1atm= 1.013x10^5 Pa

P= 1.5x 1.013x10^5 = 151950 Pa

I converted ml to m^3 (am i supposed to do this?)

26.7/ 1x10^6 = 2.67x10^-5 m^3

8.93/ 1x10^6 = 8.93x10^-5 m^3

w= -PdV

w= - (151950)x (8.93x10^-5 -2.67x10^-5)

w= -9.512 J

iii) I did the same working for this part as above

P= 2.8 x 1.013x10^5

P= 283640 Pa

w= - (283640) x (8.93x10^-5 -2.67x10^-5)

w= -17.756 J

am I doing this correct? or do I use the equation ln V2/V1 P ?
 
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i) That's free expansion; It's as if the gas is 'doing work' by pushing a massless piston in space since there's no opposing force as the gas expands.

ii & iii) It's isothermal expansion, so P varies as the volume changes. The equation W = -P ∆V doesn't work anymore and W = -∫ P dv, is used instead.

Also, yeah, it's better to convert ml to cubic meters since it's simpler to keep everything in the MKS system of units.
 
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