BRAVO! That answer is right!

The ideal gas law is

PV = n R T

where R = 8.314 J/mole-K .

What is the temperature T of half a mole of H gas in a 3-liter tank at 4 atm of pressure?

Be (very) careful with the units!  

PV = n R T

(4 atm) (3 l) = (0.5 mole) (8.314 J/mole-K) T

(0.5 mole) (8.314 J/mole-K) T = (4 atm) (3 l)

T = (4 atm) (3 l) / (0.5 mole) (8.314 J/mole-K)

T = (4 atm) (3 l) / (0.5) (8.314 J/K)

T = [ (4 atm) (3 l) / (0.5) (8.314 J) ] K

T = [ 12 l-atm / (0.5) (8.314 J) ] K

T = [ (24/ 8.314) (l-atm/J) ] K

T = [ 2.89 (l-atm/J) ] K

Now, we're down to a units-conversion question. What is a liter-atmosphere (l-atm)?

1 l = (0.10 m)³

1 l = 0.001 m³ = 1 x 10 ^{- 3} m³

1 atm = 1.013 x 10⁵ Pa

1 l-atm = (1 x 10 ^{- 3} m³)(1.013 x 10⁵ Pa)

1 l-atm = 1.013 x 10 ² m³ - Pa

Now, what's a m ³- Pa?

1 l-atm = 1.013 x 10 ² m³ - Pa [ (N/m²) / Pa ]

1 l-atm = 1.013 x 10 ² m³ [ (N/m²) ]

1 l-atm = 1.013 x 10 ² N-m

1 l-atm = 1.013 x 10 ²J

So, a liter-atmosphere is a unit of work or energy.

T = [ 2.89 (l-atm/J) ] K [(1.013 x 10 ²J)/(l-atm)]

T = [ 2.89 (1.013 x 10 ²) ] K

T = [2.89 (101.3) ] K

T = 292 K

T = 19^oC

(c) 2000, Doug Davis; all rights reserved.