Question

WORKSHEET 9 5. A 14 oz. can of hair spray contains 414.0 mL of a gas mixture containing polymers and propellant at ambient temperatures. A typical can contains 80% propellant by weight. Assume the gas is 100% propellant made of isobutane (CH). The density of isobutane is 2.51 g mL. If the amount of gas were doubled under constant pressure, what volume (in mL) would the gas occupy? Assume an ideal gas. A: 828

Answer 1

Given the can contain 414.0 mL of gas mixture. That means the volume of the gas present in can = 414.0 mL

Now the mixture is present at ambient temperature i.e at room temperature = 298 K

Given the propellent is 100% isobutane

And the given density of isobutane = 2.51 g/ml

Then mass of isobutane present = density * volume

= 2.51 g/ml * 414.0 mL

= 0.006 g

Thus moles of isobutane present = mass / molar mass

= 0.006 g / 58.12 g/mol

= 0.000104 moles

Now if we consider he gas to be an ideal gas, then pressure of the system can be calculated from the ideal gas equation, i.e

PV = nRT

where P = pressure of the system

T = temperature =  298 K

R = universal gas constant = 0.082057 L atm mol^-1K^-1,

n = number of moles of gas = 0.000104 moles

Now putting these values-

PV = nRT

P = nRT/ V

= 0.000104 moles * 0.082057 L atm mol^-1K^-1,* 298 K / 414.0 mL

= 0.000104 moles * 0.082057 L atm mol^-1K^-1,* 298 K / 0.414 L

= 0.00616 atm

Now given the amount of gas is doubled at constant pressure. that means the new number of moles of gas = 2* 0.000104 moles

THen putting the values, the new volume will be -

V = nRT/ P

= 2* 0.000104 moles * 0.082057 L atm mol^-1K^-1,* 298 K / 0.00616 atm

= 0.825 L

= 825 mL

It shows that when we double the amount of gas at constant pressure, the volume also gets doubled.