Ideal Gas Law PV = nRT. P = pressure in Pa (Absolute, not gauge) V = volume in m 3 n = moles of gas molecules n = mass/molar mass careful of: N O F Cl.

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Ideal Gas Law PV = nRT

P = pressure in Pa (Absolute, not gauge) V = volume in m 3 n = moles of gas molecules n = mass/molar mass careful of: N O F Cl Br I H R = 8.31 JK -1 (for these units) T = ABSOLUTE TEMPERATURE (in K) STP: T = K, P = x 10 5 Pa

Example – Nitrogen cylinder is at a (gauge) pressure of 90.1 psi. It has a volume of 378 liters at a temperature of 37.0 o C. What is the mass of Nitrogen in the tank? (N amu) 2967 g = 2.97 kg

Example – Nitrogen cylinder is a a (gauge) pressure of 90.1 psi. It has a volume of 378 liters at a temperature of 37.0 o C. What is the mass of Nitrogen in the tank? P = = psi P = (1.013E5 Pa/atm)(104.8 psi)/(14.7 psi/atm) = 722,193.2 Pa T = = K V = (378 liters)/(1000 liters/m 3 ) =.378 m 3 PV = nRT, n = PV/(RT) = (722,193.2 Pa)(0.378 m 3 )/((8.31 J/K)(310.15)) = mols mass = mols x molar mass mass = ( mols)( g/mol) = 2967 g = 2.97 kg

PV = nRT P = x 10 5 Pa, n = 1.0, T = 273 K, V = m 3 = 22.4 liters 22.4 liters What is the volume in liters of 1.00 mol of N 2 at 0.00 o C, and 1.00 atm? (1 atm = x 10 5 Pa)

The molecular mass of O 2 is 32 g/mol n = mass/molecular mass There are 1000 liters in a m 3 n = 34/32 = mol V =.0183 m 3 T = 23 o C = 296 K P = nRT/V 1.43 x 10 5 Pa We have 34 g of O 2 in o C. What pressure? (3)

The molecular mass of He is 4.00 g/mol n = mass/molecular mass 1000 liters = 1 m 3 n = 52.0/4.00 = 13 mol V =.212 m 3 T = PV/(nR) 422 K, 149 o C What is the temperature if 52.0 g of He occupies 212 liters at a pressure of 2.15 x 10 5 Pa?