Question

Propane gas is used as a propellant in aerosol cans.

a. Suppose a 455 mL container contains propane gas at a pressure of 1.75 atm at 25.0 ^oC. How many grams of propane are in the container?

b. If the can in part a is cooled to 5.0 ^oC and an additional 10.0 grams of propane is added to the can, what is the pressure of the propane gas? How does this pressure compare to the original pressure in part a? Whose law(s) explains this change and use the postulates of Kinetic Molecular Theory to explain the change.

c. If all the propane in the original can (as described in part a) were transferred to a balloon at STP, what volume would the propane occupy? How does this compare the original conditions in part a? Explain in terms of Kinetic Molecular Theory and gas laws.

d. The can in part a says that it can handle a pressure of 5.00 atm without bursting. What is the maximum mass of propane the can is able to hold before bursting? Whose law describes the relationship and explain the change using Kinetic Molecular Theory.

e. The can in part a says that it can handle a pressure of 5.00 atm. What is the maximum temperature to which the propane can be heated to without bursting.

Answer 1

Ans:-

a) P₁V₁/T₁ = P₂V₂/T₂     P_1,V₁ &T₁ are the given pressure, volume and temperature

P_2,V₂&T₂   are the standard pressure, volume and temperature

or, (1.75atm x 0.455L)/298K = (1atm x V₂)/273K

or, V₂ = 0.729 L

Weight of propane present in can = (44 gm x 0.729 L)/22.4 L [molecular wt. of propane = 44 gm]

= 1.432 gm

b) V₁/T₁ = V₂/T₂ at const. pressure

0.455 L/298K = V₂/278K

V₂ = 0.424 L

total wt. of propane = 1.432gm + 10gm = 11.432 gm

we know, PV = nRT

P = nRT/V

= [(11.432/44) x 0.082 L.atm.mol^-1.K^-1 x 278K]/0.424L

= 13.968 atm

here 12.218 atm more pressure create compare to part a.

Boyle's law explain the change.

Explanation:- Increasing no of moles gas at const. volume pressure must be increase.

c)

P₁V₁/T₁ = P₂V₂/T₂     P_1,V₁ &T₁ are the given pressure, volume and temperature

P_2,V₂&T₂   are the standard pressure, volume and temperature

or, (1.75atm x 0.455L)/298K = (1atm x V₂)/273K

or, V₂ = 0.729 L

volume of balloon = 0.729 L

here 0.274 L more volume create compare to part a.

Explanation:- Decrease the pressure of gas increase the volume.

d) When pressure will be greater than 5 atm then container will bursting.

PV = nRT

5 atm x 0.455 L = n x 0.082 x 298K

n = 0.093 M

maximum mass of propane contain = 0.093 M x 44 gm = 4.096 gm

Charles's and Boyle's combination law describe the relationship.

Explanation:- At const. temperature and volume mass of gas increase will be increased the pressure.

e)      PV = nRT

5 atm x 0.455L = (1.432/44) x 0.082 x T

T = 852.46 K

Maximum temperature handle by the container = 852.46 K