Chemistry; Gases

Question

A sample containing 4.80 g of O2 gas has an initial volume of 15.5 L . What is the final volume, in liters, when each of the following changes occurs in the quantity of the gas at constant pressure and temperature?

 

a) A sample of 0.400 mole of O2 is added to the 4.80 g of O2 in the container.

b) A sample of 2.20 g of O2 is removed from the 4.80 g of O2 in the container.

c) A sample of 3.10 g of O2 is added to the 4.80 g of O2 gas in the container.

Answer 1

The explanation and approach provided by Arturo is exactly correct, however, the molar mass of O2 is 32 g/mole, and not 16 g/mole, thus you will need to recalculate the moles of O2 from the given number of grams of O2.  For example, in part (a) it will be 4.8 g O2 x 1 mole O2/32 g = 0.15 moles.  This added to the o.40 moles = 0.55 moles.  Do the same for parts (b) and (c).

Answer 2

Edited answer after correcting for molecular weight of diatomic oxygen:

PV = nRT

 

V = nRT/P

 

If P and T are constant, then RT/P is constant, and V is proportional to n (number of moles).  Find out how the number of moles changes in each case.

 

V' = (n'/n)V

 

(a)

 

n₁ = 4.80/32.0 moles = 0.150 moles

n₂ = n₁ + 0.400 = 0.150 + 0.400 = 0.550 moles

 

V₂ = (n₂/n₁)V₁ = (0.550/0.150)(15.5 L) = 56.8 L

 

(b)

 

n₂ = n₁ - 2.20/32.00 moles = 0.150 - 0.1375 = 0.0813 moles

 

V₂ = (n₂/n₁)V₁ = (0.0813/0.150)(15.5 L) = 8.40 L

 

(c)

 

n₂ = 3.10/32.00 + n₁ = 0.0969 + 0.150 = 0.247 moles

 

V₂ = (n₂/n₁)V₁ = (0.247/0.150)(15.5 L) = 25.5 L