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

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

Step-by-step 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

`(PV)/(Rn) = T\\\\since \ temperature\ T \ is \ constant;\\\\(P_1V)/(Rn_1) = (P_2V)/(Rn_2)\\\\(P_1)/(n_1) = (P_2)/(n_2) \\\\n_2 = (n_1P_2)/(P_1) \\\\n_2 = (1.39 * 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.