Thermodynamics Ideal Gas Law problem (pressure problem)

AI Thread Summary
The discussion revolves around a thermodynamics problem involving the Ideal Gas Law, where a gas initially at standard temperature and pressure (STP) is subjected to a new pressure and temperature. The initial volume is 2.00 m³, with the gas being compressed under a pressure of 5.00 atm and heated to 39.0°C. The calculations initially used an incorrect pressure of 6 atm instead of the specified 5 atm, leading to confusion in the solution. The correct application of the Ideal Gas Law shows that the final volume, after adjustments, is approximately 0.3809 m³. The importance of using Kelvin for temperature and understanding pressure ratios without conversion to metric units is emphasized.
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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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