Question

A gaseous mixture in a 3 litre vessel contains 0.1 mol H2, 0.3 mol He at 27 °C and 2 bar.
Find the partial pressures of H, and He.​

Answer 1

Explanation:

$\huge\purple{\mathbb{Question: }}$

A gaseous mixture in a 3 litre vessel contains 0.1 mol H2, 0.3 mol He at 27 °C and 2 bar.

Find the partial pressures of H, and He.

$\huge\red{\mathbb{answer: }}$

$Moles of N2= 0.3$

$Moles of O2= 0.1$

$Moles of He = 0.1$

$Mole fraction of O2=$

$no2 \div nn2 + no2 + nhe \\ </p><p>\\ = 0.1 \div 0.3 + 0.1 + 0.2 \\ = 02$

pressure = 0.2 ×720 = 144

Answer 2

GIVEN :–

• A gaseous mixture in a 3 litre vessel contains 0.1 mol H₂ , 0.3 mol He at 27 °C and 2 bar.

TO FIND :–

• Partial pressures of H₂ , and He.

SOLUTION :–

• According to the question –

$\\\bf \to Vessel \: \: contains \: \: 0.1 \: \: mol \: \: of \: \: H_{2}$

$\\\bf \to Vessel \: \: contains \: \: 0.3\: \: mol \: \: of \: \: He$

• Now let's calculate mole fraction of gases –

$\\\bf \implies Mol \: \: fraction \: \: of \: \: H_{2}= \dfrac{n_{H_{2}}}{n_{H_{2}} + n_{He}}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: H_{2}= \dfrac{0.1}{0.1 + 0.3}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: H_{2}= \dfrac{0.1}{0.4}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: H_{2}= \dfrac{1}{4}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: H_{2}=0.25$

• Similarly –

$\\\bf \implies Mol \: \: fraction \: \: of \: \: He= \dfrac{n_{He}}{n_{H_{2}} + n_{He}}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: He= \dfrac{0.3}{0.1 + 0.3}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: He= \dfrac{0.3}{0.4}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: He= \dfrac{3}{4}$

$\\\bf \implies Mol \: \: fraction \: \: of \: \: He=0.75$

• Total pressure = 2 bar

• So that –

$\\\implies \large \red{\boxed{ \bf Partial \: \: pressure \: \: of \: \:H_{2} = Mol \: \: fraction \: \: of \: \: H_{2} \times Total \: \: pressure}}$

$\\\implies \bf Partial \: \: pressure \: \: of \: \:H_{2} = 0.25 \times 2$

$\\\implies \bf Partial \: \: pressure \: \: of \: \:H_{2} = 0.5$

$\\\implies \large\pink{ \boxed{ \bf Partial \: \: pressure \: \: of \: \:H_{2} = 0.5 \: bar}}$

• And –

$\\\implies \large \red{\boxed{ \bf Partial \: \: pressure \: \: of \: \:He = Mol \: \: fraction \: \: of \: \: He \times Total \: \: pressure}}$

$\\\implies \bf Partial \: \: pressure \: \: of \: \:He= 0.75 \times 2$

$\\\implies \bf Partial \: \: pressure \: \: of \: \:He= 1.5$

$\\\implies \large\pink{ \boxed{ \bf Partial \: \: pressure \: \: of \: \:He = 1.5 \: bar}}$

☞ Henceforth , Partial pressure of H & He is 0.5 bar & 1.5 bar respectively.