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