In: Chemistry
A 6.79-L cylinder contains 2.00 mol of gas A and 3.96 mol of gas B, at a temperature of 31.5 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.
volume = V = 6.79 L
total moles = A + B = 2 + 3.96 = 5.96 mol
mole fraction of A = 2 / 5.96 = 0.336
mole fraction of B = 3.96 / 5.96 = 0.664
T = 31.5 + 273 = 304.5 K
PV = n RT
P x 6.79 = 5.96 x 0.0821 x 304.5
P = 21.94 atm
total pressure = 21.94 atm
partial pressure of A = A mole fraction x total pressure
= 0.336 x 21.94
= 7.37 atm
partial pressure of A= 7.37 atm
partial pressure of B = B mole fraction x total pressure
= 0.664 x 21.94
= 14.57 atm
partial pressure of A= 14.57 atm