Question

A sample of argon measuring 2.75 L at 29.0 degrees C and 764 torr was bubbled through and collected over water. The pressure of the new system is 789 torr at 15.0 degrees C. What is the volume occupied by the gas? The vapor pressure of water at 15.0degrees C is 13 torr.

Answer 1

Given Conditions,

volume of the gas = 2.75 L

temperature = 29 0C = 29+273 = 302K

presuure = 764 torr = 764/760 = 1.0053 atm

initial number of moles of Ar n = PV/RT according to ideal gas equation.

n = 1.0053 * 2.75/(0.0821 * 302)

n = 0.1115 moles

pressure of new system = 789 torr

pressure of Ar gas = pressure of the system - pressure of water

Pressure of Ar gas = 789 - 13 = 776 torr 776/760 = 1.0131 atm

temperature = 15 0C = 15+273 = 288K

number of moles = 0.1115 moles

again from ideal gas equation,

V = nRT/P

V = volume of the Ar gas = 0.1115*0.0821*288/1.0131

V = 2.6023 L