Question

1. Water can be split into hydrogen and oxygen under suitable conditions.
   equation representing the change is :
   2H₂O(l) --------------- 2H₂(g) + O₂(g).

(i) If a given experiment result in 2500 cm³ of hydrogen being produced, what value of oxygen is liberated at the same time under the same conditions of temperature and pressure?

(ii) The 2500 cm³ of hydrogen is subjected to 2 _1/2 times increases in pressure (temperature remaining constant ). What volume will the hydrogen now occupy ?

(iii) Taking the volume of hydrogen calculated in (d) (ii), what change must be made in the kelvin (absolute) temprature to return the volume to 2500 cm³ (pressure remaining constant)?

Answer 1

(i) 2 Vol. of H2 = 1 vol. of O2
2500 of H2 = 1/2 x 2500
=1250 cm3 of O2

(ii) P1 = 1 Atm.
P2 = 2.5 Atm.
V1 = 2500
V2 = ?
P1 V1 = P2 V2
1 x 2500 = 2.5 x V2
V2 = 2500 x 10/25
V2 = 1000 cm3

(iii) (If volume of H2 = 1000 cm3 say, temp. is T1, V2 = 2500 New temp. T2
V1 / T1 = V2 / T2
1000/T1 = 2500/T2
or
T2/T1= 2500/1000
= 2.5
T2 = 2.5 T1
It must be 2.5 times of original temperature.