Expansion of a gas at constant temperature and pressure

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SUMMARY

The discussion focuses on calculating the work done by a gas during isothermal expansion at constant temperature and pressure. The work done against a vacuum is zero, while the calculations for work done against constant pressures of 1.5 atm and 2.8 atm yield -9.512 J and -17.756 J, respectively. The correct approach involves using the equation w = -PdV, and it is essential to convert volumes from milliliters to cubic meters for accurate results. The discussion emphasizes the importance of using the MKS system of units for consistency in calculations.

PREREQUISITES
  • Understanding of isothermal processes in thermodynamics
  • Familiarity with the ideal gas law
  • Knowledge of unit conversions, particularly from mL to m³
  • Proficiency in calculus for integrating pressure over volume
NEXT STEPS
  • Study the principles of isothermal expansion in thermodynamics
  • Learn about the integration of pressure in work calculations for variable pressures
  • Explore the ideal gas law and its applications in real-world scenarios
  • Review unit conversion techniques, especially within the MKS system
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Students studying thermodynamics, physics enthusiasts, and professionals involved in gas dynamics or engineering applications related to gas behavior during expansion.

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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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