Answer:
See Explanation
Explanation:
These problems are based upon the Ideal Gas Law, PV = nRT. Variables are gas phase variables and include ...
Pressure (P)
Volume (V)
Mass (moles (n))
R = Gas Constant = 0.08206L·atm/mol·K
Temperature (T)
Ideal Gas Law => PV = nRT
Note => when using the gas constant, R, convert all dimensional units to match those of L·atm/mol·K.
Problem 1 => Given ...
P = nRT/V
V = 250 ml = 0.250 L
n = 3.2 moles
R = 0.08206L·atm/mol·K
T = 45°C = (45 + 273)K = 318K
P = nRT/V = (3.2mol)(0.08206L·atm/mol·K)(318K)/(0.250L) = 334atm = 334atm(760mmHg/atm) = 253,853mmHg ≅2.5 x 10⁵mmHg (2 sig. figs.)
Problem 2 = Given ...
P = 5.05atm
V = nRT/P
n = 4.25 moles
R = 0.08206L·atm/mol·K
T = 27°C = (27 + 273)K = 300K
V = (4.25mol)(0.08206L·atm/mol·K)(300K)/(5.05atm) = 20.7 Liters O₂(g)
Problem 3 = Given ...
P = 302 KPa = (302KPa)(0.01atm/KPa) = 3.02 atm
V = 4.5 Liters
n = PV/RT
R = 0.08206L·atm/mol·K
T = 15°C = (15 + 273)K = 288K
n = (3.02atm)(4.5L)/(0.08206L·atm/mol·K)(288k) = 0.58 mole O₂ = 0.58 mole x 32 g/mole = 18.4 grams O₂
Problem 4 = Given ...
P = 810mmHg = 810mmHg(1atm/760mmHg) = 1.07atm
V = 2.0 Liters
n = 0.56 mole N₂(g)
R = 0.08206L·atm/mol·K
T = PV/nR = (1.07atm)(2.0L)/(0.56mol)(0.08206L·atm/mol·K) = 46.6K = (46.6 - 273)°C = -226°C
Problem 5 = Given ...
P = 3.2 atm
V = 55 Liters
n = PV/RT
R = 0.08206L·atm/mol·K
T = 17°C = (17 + 273)K = 290K
n = (3.2atm)(55L)/(0.08206L·atm/mol·K)(290K) = 7.4 moles N₂(g)
