Question

${\text{100 ml of P}}{{\text{H}}_3}$, when decomposed produces phosphorus and hydrogen. The change in volume is:
A.$50\,{\text{ml}}$ increase
B.${\text{500 ml}}$ decrease
C.${\text{900 ml}}$ decrease
D.None of the above

Answer 1

Hint:Gay-Lussac, s law often recognized as the pressure law which is defined as the pressure of an enclosed gas is directly proportional to the temperature. Gay-Lussac, s shows the relationship between the pressure and temperature of a fixed mass of gas kept at a constant volume. This law is also known as the pressure law.
$\dfrac{{{{\text{P}}_1}}}{{{{\text{T}}_1}}}\, = \,\dfrac{{{{\text{P}}_2}}}{{{{\text{T}}_2}}}$
Law of combining volumes: The law of combining volume states that, when the gases react together to form other gases and when all the volumes are measured at the same temperature and pressure. The ratio between the volume of the reactant gases and gaseous products can be expressed in a simple whole number.

Complete step by step answer:
Let us look at the formation of water molecules.
${\text{2}}{{\text{H}}_2}\, + \,{{\text{O}}_2}\, \to \,{\text{2}}\,{{\text{H}}_2}{\text{O}}$
 Gay-Lussac, s found that $2$ volumes of hydrogen and $1$ volume of oxygen react to form $2$ volumes of water . Based on Gay-Lussac, s results , Amedeo Avogadro theorized that , at the same temperature and pressure , equal volumes of gas contain equal numbers of molecules.
So the thermal decomposition of phosphine (${\text{P}}{{\text{H}}_3}$) , into phosphorus and molecular hydrogen is a first order reaction and given by
$4\,{\text{P}}{{\text{H}}_3}\,\left( g \right)\, \to \,{{\text{P}}_4}\,\left( g \right)\,\, + \,6\,{{\text{H}}_2}\,\left( g \right)$
By applying Gay-Lussac, s law of combining volumes , when temperature and pressure remains the same , ${\text{4 ml of P}}{{\text{H}}_3}$will give $6\,{\text{ml of }}{{\text{H}}_2}$ .
$
  4\,{\text{ml }}\,{\text{of}}\,{\text{P}}{{\text{H}}_3}\, \to \,\,6\,{\text{ml of }}\,{{\text{H}}_2} \\
  1\,{\text{ml }}\,{\text{of }}\,{\text{P}}{{\text{H}}_3}\, \to \,\dfrac{6}{4}\,{\text{ml }}\,{\text{of }}\,{{\text{H}}_2}\,,\,{\text{That is 1ml of}}\,{\text{P}}{{\text{H}}_{_3}}\, \to \,\dfrac{3}{2}\,{\text{ml of}}\,{{\text{H}}_2} \\
  \therefore \,\,x\,{\text{ml of}}\,{\text{P}}{{\text{H}}_3}\, \to \,x\, \times \,\dfrac{3}{2}\,{\text{ml of}}\,{{\text{H}}_2} \\
 $
So in the case of \[{\text{100 ml of P}}{{\text{H}}_3}\]
$
  100\,{\text{ml}}\,\,{\text{of}}\,{\text{P}}{{\text{H}}_3}\, \to \,100 \times \dfrac{3}{2}\,\,{\text{ml}}\,\,{\text{of}}\,{{\text{H}}_2} \\
  100\,{\text{ml of}}\,{\text{P}}{{\text{H}}_3}\, \to \,150\,{\text{ml of}}\,{{\text{H}}_2} \\
 $
The initial volume of ${\text{P}}{{\text{H}}_3}{\text{ is 100 ml}}$, on decomposition give $150{\text{ ml of }}{{\text{H}}_2}$. So here the volume is increased by $50\,{\text{ml}}$.
So here the change in volume is $50\,{\text{ml}}$.

Hence, the correct answer is option A.

Note:
The combined gas law is a useful starting point for all cases with gases, even if one of the variables is a constant. The variable that is constant can easily be eliminated from the combined gas law equation, reducing it back to Boyle s or Charles law.