Question

. If the volume of 2 mol oxygen gas O2 is 3 L, the volume of 3 mol chlorine gas Cl2 at
the same condition is equal.
a. 3.0 L
b. 4.5 L
c. 2.0 L
d. 9.0 L

Answer 1

Answer:

P V= nRT

P×3= 2×R×T -(1)

P×V=3×R×T -(2)

3/V = 2/3

V= 9/2

V=4.5 L

Answer 2

Answer :-

The required volume occupied by Chlorine gas is 4.5 Litres [Option.b]

Explanation :-

We have :-

→ Moles of O₂ (n₁) = 2 moles

→ Volume occuiped by O₂ (V₁) = 3 L

→ Moles of Cl₂ (n₂) = 3 mol

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So, by putting values in the Ideal Gas Equation, we get the two equations as :-

⇒ P × V₁ = n₁ × R × T

⇒ 3P = 2RT ---(1)

⇒ P × V₂ = n₂ × R × T

⇒ PV₂ = 3RT ---(2)

On dividing eq(1) by eq(2), we get :-

⇒ 3P/PV₂ = 2RT/3RT

⇒ 3/V₂ = 2/3

⇒ 2V₂ = 9

⇒ V₂ = 9/2

⇒ V₂ = 4.5 L