The combined gas law P1 V1 P2 V2 = T1 T2 Units: ____________ ____________ = T1 T2 Units: P any unit provided they match V any unit provided they match T must be in Kelvin
Ideal gas law PV = nRT P = absolute pressure V = volume n = # of mols R = universal gas constant T = absolute temperature
units: T must be in Kelvin Pressure Volume R Pa m3 8.31 J/mol·K n = m/M where: m = mass in grams and M = molar mass in g/mol
First Law of Thermodynamics U = Q + W Where : U change in internal energy Q heat added or removed W work done
sign notation: W positive if work is done on the gas negative if work is done by the gas Q positive if heat is added negative if heat is removed U positive if temperature increases negative if temperature decreases
where ( T = Tf - Ti ) in all cases EQUATIONS W = - ∫P V where ( V = Vf - Vi ) Q = n cp T - if pressure is constant or Q = n cv T - if volume is constant U = 3/2 nR T where ( T = Tf - Ti ) in all cases
CONSTANTS cp = 5/2 R ( 20.775 J/K ) cv = 3/2 R ( 12.465 J/K ) R = 8.31 J/mol·K
UNITS U JOULES W JOULES Q JOULES Therefore: Pressure must be in Pa Volume must be in m3 R is 8.31 J/molK
The temperature of 3 kg of the monatomic gas krypton is raised from -20oC to 80oC. If this is done at constant volume, compute the heat added, the work done, and the change in internal energy. The molecular weight of Kr is 83.7 g/mol. 3.0 kg 1000 g 1 mol x x = 35.84 mols = n 1 kg 83.7 g T = 80 – -20 = 100 C° = 100K U = 3/2 nR T = 3/2 (35.84)(8.31)(100) = 44,700 J Q = n cv T = (35.84)(12.47)(100) = 44,700 J U = Q + W ▬► 44,700 = 44,700 + W W = 0 J* * W could have also been determined by W = -∫PV (since V = 0, W = 0)
Repeat the previous problem if the heating process were at constant pressure instead of constant volume. T and the number of mols are still the same, thus U is still the same U = 44,700 J NOW: Q = n cp T = (35.84)(20.78)(100) = 74,480 J U = Q + W ▬► 44,700 = 74,480 + W W = -29780 J Work is done by the gas since it is negative