What Is the Work Done by the Gas and the Final Gas Temperature?

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SUMMARY

The discussion centers on the thermodynamic analysis of an ideal gas in a piston-cylinder device, specifically addressing the work done by the gas and the final gas temperature. The work done by the gas is calculated as -132.54 J, indicating that the gas does work on the piston, while the final gas pressure remains at 1.5x105 Pa. The final gas temperature is determined to be 450 K, with a total heat transfer of 464.26 J during the process. The specific heat at constant volume is confirmed as 718 J/kg·K, and the gas constant used is 287 J/kg·K.

PREREQUISITES
  • Understanding of ideal gas laws and properties
  • Knowledge of thermodynamic processes, specifically constant volume and constant pressure
  • Familiarity with the concepts of work done by gases in piston systems
  • Basic principles of energy conservation in thermodynamic systems
NEXT STEPS
  • Study the derivation and application of the ideal gas law
  • Learn about the relationship between specific heats and internal energy
  • Explore the concept of enthalpy and its relevance in constant pressure processes
  • Investigate the implications of negative work done in thermodynamic systems
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in the study of gas behavior in piston-cylinder systems.

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Homework Statement


An ideal gas at 300 K and 1.5x10^5 Pa is contained in a piston cylinder device. The piston is 0.1m above the base of the cylinder, has a diameter of 0.15 m and is allowed to move freely. Heat is transferred to the cylinder causing the piston to move a distance of 0.05 m. The gas specific heat at constant volume is 718 J/kg·K and the gas constant is 287 J/kg·K.
(a) What is the work done by the gas and the gas pressure at the end of the process? (b) Calculate the final gas temperature and the amount of heat transferred to the gas during the whole
process. (Ans: (a) -132.54J, 1.5x10^5 Pa, (b) 450K, 464.26 J )

Homework Equations


The Attempt at a Solution


a)
The pressure is constant = 1.5x105 Pa I have no idea why...
WD by gas = (1.5x10^5)(0.1-0.05)pi(0.075^2)=132.54J <---- It should be positive right ? The answer shows negative.
b)
T=300(0.15A)/(0.1A)=450K , A is the area.
OK the main problem starts here. After a few tries i got
n=PV/(T287) <-------------so this one i should take 287 instead of the usual 8.31? how come ? I know the qn says gas contant is 287 but why did it change ?
deltaU=(fR/2)(ndeltaT) <---------- Is this eqn right ? f is the degree of freedom since we aren't told what type of gas this is.
I took fR/2 as the specific heat at constant vol( correct?)
and obtain deltaU=331.57
For energy conservation the heat supplied must cause the increase in U by 331.57 and must also provide for the WD by gas. Q=331.57+132.54=464.11 Actual answer is 464.26.
I cannot grasp this properly. This is a constant pressure process. So why do we use constant vol specific heat?
 
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Work should be positive.

Pressure is constant because problem says the piston is allowed to move freely. The portion above the piston is open to the atmosphere.

One uses constant pressure specific heat when one deals with enthalpy. You are dealing with internal energy so constant volume specific heat is used.
 

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