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If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under the same conditions?

Sagot :

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

[tex]n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1[/tex]

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

[tex]n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol[/tex]

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L[/tex]

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

The volume occupied by 3.0 g of He is mathematically given as

V2=16.8L

What volume does the 3.0 g of He occupy?

Question Parameters:

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of H

Generally, the equation for the gas equation  is mathematically given as

PV=nRT

When the reacting mass is 3g

n2=3/4

n2=0.75

Therefore

V1/n1=V2/n2

Where

n1=1

[tex]V2=\frac{22.4*0.75}{1}[/tex]

V2=16.8L

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