Question

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

Answer 1

The volume occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19°C is approximately 3.23 L.

Step-by-step explanation:

To find the volume occupied by 3.760 g of oxygen gas, we first need to convert the mass of oxygen to moles. The molar mass of oxygen (O2) is 32 g/mol, so we divide the mass by the molar mass to get the number of moles. Next, we convert the temperature from Celsius to Kelvin by adding 273.15. Then, we can use the ideal gas law equation PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. Rearranging the equation to solve for V, we get V = (nRT)/P. Plugging in the values, we get V = (moles)(8.314 J/mol·K)(temperature)/pressure. Calculating the volume gives us the answer.

Plugging in the values, we get V = (moles)(8.314 J/mol·K)(temperature)/pressure. Calculating the volume gives us the answer.

V = (0.117 mol)(8.314 J/mol·K)(292.15 K)/(88.4 kPa) = 3.23 L

Answer 2

3.2 L

Step-by-step 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