Water decomposes to O₂ and H₂ under suitable conditions as represented by the equation below:

2H₂O ⟶ 2H₂ + O₂

(a) If 2500 cm³ of H₂ is produced, what volume of O₂ is liberated at the same time and under the same conditions of temperature and pressure?

(b) The 2500 cm³ of H₂ is subjected to $2\dfrac{1}{2}$ times increase in pressure (temp. remaining constant). What volume will H₂ now occupy?

(c) Taking the volume of H₂ calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm³ pressure remaining constant.

$\begin{matrix} 2\text{H}$2\text{O} & \longrightarrow & 2\text{H}2 & + &\text{O} \ 2\text{ Vol.} & & 2\text{ Vol.}& & 1\text{ Vol.} \end{matrix}

2 Vol. of water gives 2 Vol. of H₂ and 1 Vol. of O₂

∴ If 2500 cm³ of H₂ is produced, volume of O₂ produced = $\dfrac{2500}{2}$ = 1250 cm³

(b) V₁ = 2500 cm³

P₁ = 1 atm = 760 mm

T₁ = T

T₂ = T

P₂ = [760 x 2 $\dfrac{1}{2}$] + [760] = 760 [$\dfrac{5}{2}$ + 1] = 760 x $\dfrac{7}{2}$ = 2660 mm

V₂ = ?

Using formula:

$\dfrac{\text{P}$1\text{V}1}{\text{T}1} = $\dfrac{\text{P}2\text{V}$2}{\text{T}2}

$\dfrac{760 \times 2500}{\text{T}}$ = $2660 \times \dfrac{\text{V}_2}{\text{T}}$

V₂ = $\dfrac{760 \times 2500 }{2660}$ = $\dfrac{5000}{7}$

(c) V₁ = $\dfrac{5000}{7}$ = 714.29 cm³

P₁ = P₂ = P

T₁ = T

V₂ = 2500 cm³

T₂ = ?

Using formula:

$\dfrac{\text{P}$1\text{V}1}{\text{T}1} = $\dfrac{\text{P}2\text{V}$2}{\text{T}2}

$\dfrac{\text{P} \times 714.29}{\text{T}}$ = $\dfrac{\text{P} \times 2500}{\text{T}_2}$

T₂ = $\dfrac{2500 }{714.29}$ x T

T₂ = 3.5 x T

Hence, T₂ = 3.5 times T or temperature should be increased by 3.5 times