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 Tof 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 m

^{3}= 1 x 10^{ - 3}m^{3}1 atm = 1.013 x 10

^{5}Pa1 l-atm = (1 x 10

^{ - 3}m^{3})(1.013 x 10^{5}Pa)1 l-atm = 1.013 x 10

^{ 2}m^{3 }- PaNow, what's a m

^{ 3}- Pa?1 l-atm = 1.013 x 10 ^{ 2}m^{3 }- Pa [ (N/m^{2}) / Pa ]1 l-atm = 1.013 x 10

^{ 2}m^{3 }[ (N/m^{2}) ]1 l-atm = 1.013 x 10

^{ 2}N-m1 l-atm = 1.013 x 10

^{ 2}JSo, a liter-atmosphere is a unit of

workorenergy.T = [ 2.89 (l-atm/J) ] K [(1.013 x 10 ^{ 2}J)/(l-atm)]T = [ 2.89 (1.013 x 10

^{ 2}) ] KT = [2.89 (101.3) ] K

T = 292 K

T = 19^{o}C

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