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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
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Graham’s Law of Diffusion calculation

 Graham’s law states that any gas's diffusion rate at constant temperature and pressure is directly proportional to the square root of its relative molecular mass or vapour density.

R1/R2 = √m2/√m1                    or t2/t1 = √m2/√m1

Example Questions

1.       If 200cm3 of hydrogen diffused through a porous pot in 30 seconds. How long will it take 400cm3 of oxygen to diffuse through the same pot? (H = 1, O = 16)

Solution

Before solving the problem, the first thing is to find how long it will take an equal volume (400cm3) of hydrogen to diffuse through the pot. This is because we are comparing oxygen and hydrogen.

Since 200cm3 of hydrogen = 30s

400cm3 will = 30/200 X 400 = 60s

        t2/t1 = √m2/√m1

        .t1 = 60s

         .t2 = ??

         M1 = 2

          M2 = 32

               60/.t2 = √2/√32   

            .t2 2  = 60x60 x 16

                   = 57600

            .t2 = √57600 = 240sec

2.       Under the same conditions of temperature and pressure hydrogen diffuses 2 times as fast as a gas, X. Calculate the relative molecular mass of X (relative molecular = 2)

R of hydrogen = 2

R  of X  = 1

M of H =  2

1/2 = √2/√m

Square both sides to remove the square root

1/4 = 2/mx                       Mx = 4x2 = 8

Relative molecular mass of X = 8

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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
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