Seeming simple Thermodynamics problem

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SUMMARY

This discussion addresses a thermodynamics problem involving a balloon inflated with 5 liters of monatomic gas at 0 degrees Celsius and heated to 60 degrees Celsius at 1 atm pressure. The calculations reveal that the heat absorbed by the gas at constant volume is approximately 166 Joules, while the work done by the gas during free expansion at constant pressure is approximately 111 Joules. Key equations utilized include the ideal gas law (PV=nRT) and the internal energy change formulas (ΔU=nCvΔT and ΔU=nCpΔT).

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Familiarity with thermodynamic concepts of internal energy (ΔU)
  • Knowledge of specific heat capacities (Cp and Cv) for monatomic gases
  • Basic proficiency in unit conversions (liters to cubic meters, Celsius to Kelvin)
NEXT STEPS
  • Explore the implications of the first law of thermodynamics (ΔU=Q+W)
  • Learn about boundary work in thermodynamic systems
  • Investigate the behavior of monatomic gases under varying conditions
  • Study the differences between constant volume and constant pressure processes
USEFUL FOR

This discussion is beneficial for students and professionals in physics and engineering, particularly those focusing on thermodynamics, as well as anyone interested in understanding gas behavior under thermal conditions.

StickNinja
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A balloon is inflated with 5 liters of monatomic gas at 0 degrees Celsius and heated to 60 degrees Celsius at 1 atm. pressure.
R=8.31J/mol K, Cp=5R/2, and Cv = 3R/2

1. If the balloon does not expand, the heat absorbed by the gas would be what?
2. If the balloon expands freely at constant pressure, the work done by the gas would be what?

...Took me a while, but I figured it out (no mass is entering the system). So here is the solution in case others want to know.

For #1 PV=nRT
P=1 atm=1.01*10^5 Pascals
V=5 liters=.005 m^3
n=Unknown
R=8.31 J/mol K
T=273K, Given

Solving for n, we get .222601 mols

ΔU=nCvΔT --constant volume
ΔU=n*3R/2*60= 166J roughlyFor #2

ΔU=nCpΔT
ΔU=277.47
ΔU=Q+W
W=111J
 
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You'll have to take into account the mass flowing into the system. There may also be boundary work involved.
 

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