Question

In: Chemistry

A 6.79-L cylinder contains 2.00 mol of gas A and 3.96 mol of gas B, at...

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.

Solutions

Expert Solution

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


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