Under the same conditions of temperature and pressure, how many times faster does protium … — JAMB UTME

Under the same conditions of temperature and pressure, how many times faster does protium gas ( $\text{H}_2$ ) diffuse through a porous plug compared to tritium gas ( $\text{T}_2$ )? [Relative atomic masses: protium, $^{1}\text{H} = 1.0$ ; tritium, $^{3}\text{H} = 3.0$ ]

$1.73$ Cevap
B
$3.00$
C
$0.58$
D
$9.00$

Cevap

Protium gas diffuses approximately

1.73

times faster than tritium gas.

The relative rate of diffusion of protium gas (

\text{H}_2

) relative to tritium gas (

\text{T}_2

) is governed by Graham's Law, which states that

R_1 / R_2 = \sqrt{M_2 / M_1}

. Given

M(\text{H}_2) = 2\text{ g mol}^{-1}

and

M(\text{T}_2) = 6\text{ g mol}^{-1}

, the ratio is

\sqrt{6 / 2} = \sqrt{3} \approx 1.73

. Thus, protium gas diffuses

1.73

times faster.

Adım Adım Çözüm

Calculate the molar mass of diatomic protium gas (

\text{H}_2

) and tritium gas (

\text{T}_2

M(\text{H}_2) = 2 \times 1.0 = 2.0\text{ g mol}^{-1}

and

M(\text{T}_2) = 2 \times 3.0 = 6.0\text{ g mol}^{-1}

Graham's law of diffusion requires the molar masses of the gaseous species.

Apply Graham's Law of diffusion:

\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{M_{\text{T}_2}}{M_{\text{H}_2}}}

\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{6.0}{2.0}} = \sqrt{3.0}

The rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass.

Evaluate the square root to determine the ratio.

\sqrt{3.0} \approx 1.73

This yields the relative diffusion rate of protium gas compared to tritium gas.

Anahtar Kavram

Graham's Law of Diffusion applied to Hydrogen Isotopes

Tahmini Süre:1m 15s

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