Thermodynamics Ideal Gas Law problem (pressure problem)

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SUMMARY

The discussion revolves around solving a thermodynamics problem involving the Ideal Gas Law, specifically calculating the volume of a gas under varying pressure and temperature conditions. The initial volume (V1) is 2.00 m³ at standard temperature and pressure (STP), with a final pressure (P2) of 5.00 atm and a final temperature (T2) of 39.0°C. The correct calculation yields a final volume (V2) of 0.3809 m³, confirming the inverse relationship between pressure and volume. Key points include the importance of using Kelvin for temperature and the correct interpretation of pressure units.

PREREQUISITES
  • Understanding of the Ideal Gas Law and its applications
  • Knowledge of unit conversions between Celsius, Kelvin, and atmospheric pressure
  • Familiarity with basic thermodynamic principles, particularly the relationship between pressure and volume
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the Ideal Gas Law and its derivations in detail
  • Learn about the significance of STP in thermodynamic calculations
  • Explore the concept of pressure-volume relationships in gases
  • Practice solving similar thermodynamic problems using different gas laws
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling gas law problems, as well as educators looking for examples of practical applications of the Ideal Gas Law.

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Homework Statement


If 2.00 m³ of a gas initially at STP is placed under a pressure of 5.00 atm, the temperature of the gas rises to 39.0°C. What is the volume?

STP means it's at 0°C and 1 atm of Pressure, correct?
V1 = 2.00 m³
V2 = ?
T1 = 0°C = 273°K (My main confusion was units: °C vs. °K and atm vs. Pa)
T2 = 39°C = 312°K
P1 = 1 atm = 101.3 kPa
P2 = 6 atm = 607.8 kPa
R = 8.135

Homework Equations


P1*V1 = nR*T1 (nR is a constant) (I don't think I need to use this, but this one translates into the next equation)
(P1*V1)/(T1) = (P2*V2)/(T2)

The Attempt at a Solution


101.3*2/273 = 607.8*V2/312
.742 = 1.948*V2
V2 = .3809 m³

This makes sense to me, because pressure is inversely proportional to Volume. As pressure increases, Volume decreases. Am I correct in my work and thinking?
 
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Looks good to me. Kelvin is the correct choice of unit.
 
You are using 6 atm. for the new pressure. It says that it is placed under a pressure of 5 atm. not an additional 5 atm. So you have to use 5 atm as the new pressure, not 6 atm. Also, you don't have to convert pressure to metric since it is just the pressure ratios you use. Other than that, as Winzer says, your method is fine.

AM
 

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