# Thermodynamics Ideal Gas Law problem (pressure problem)

1. Jan 9, 2010

### Ling_Ling

1. The problem statement, all variables and given/known data
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

2. Relevant 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)

3. 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?

2. Jan 9, 2010

### Winzer

Looks good to me. Kelvin is the correct choice of unit.

3. Jan 9, 2010

### Andrew Mason

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