Question

A rigid container has 44.5 grams of oxygen gas at room temperature and a pressure of 2.3 atm. How many grams of oxygen should the container have for the pressure to be 7.8 atm?

Answer 1

Answer:

The mass of oxygen the container must have is 150.85 g.

Explanation:

Given;

mass of the oxygen, m₁ = 44.5 g

initial pressure of the gas, P₁ = 2.3 atm

final pressure of the gas, P₂ = 7.8 atm

Atomic mass of oxygen gas, = O₂ = 16 x 2 = 32 g

initial number of moles of oxygen in the container, n₁ = 44.5/32 = 1.39

let the final number of moles of oxygen = n₂

Apply ideal gas equation;

PV = nRT

$\frac{PV}{Rn} = T\\\\since \ temperature\ T \ is \ constant;\\\\\frac{P_1V}{Rn_1} = \frac{P_2V}{Rn_2}\\\\\frac{P_1}{n_1} = \frac{P_2}{n_2} \\\\n_2 = \frac{n_1P_2}{P_1} \\\\n_2 = \frac{1.39 \times 7.8}{2.3} \\\\n_2 = 4.714 \ moles$

The mass of the oxygen in grams is calculated as;

m₂ = 4.714 x 32g

m₂ = 150.85 g

Therefore, the mass of oxygen the container must have is 150.85 g.