### 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 = 19oC