What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4
kPa and a temperature of 19°C? Assume the oxygen is ideal. Use "8.314"
for R. Hint: Convert 3.760 g of oxygen (O2) to moles first, and remember
that temperature needs to be in Kelvin.
O 6.45 L
0.210 L
O 3.23 L
0 0,420 L

3.2 L

Explanation:

Given data:

Mass of oxygen = 3.760 g

Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)

Temperature = 19°C (19+273.15 = 292.15 K)

R = 8.314 Nm K⁻¹ mol⁻¹

Volume occupied = ?

Solution:

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 3.760 g/ 32 g/mol

Number of moles = 0.12 mol

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant

T = temperature in kelvin

V = nRT/P

V = 0.12 mol ×  8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²

V = 291.472 Nm /88400 Nm⁻²

V = 0.0032 m³

m³ to L:

V = 0.0032×1000 = 3.2 L