Question

What will be the change in volume when 100mL $P{H_3}$ decomposed to solid phosphors and ${H_2}$ gas.
(A) Increase in 50ml
(B) Decrease in 50ml
(C) Increase in 150ml
(D) Decrease in 200ml

Answer 1

For any gas, the volume covered by the gas depends upon its amount in moles. We can say that 1 mole of a gas requires a volume of 22.4L at STP.

Complete step by step answer:
So, first see what will happen when 100mL of $P{H_3}$ is decomposed to solid phosphorus, So to find out this we need to see the equation for the decomposition reaction.
Therefore, the equation for the decomposition reaction would be
$4P{H_{3(g)}} \to {P_{4(s)}} + 6{H_{2(g)}}$

In the above reaction we can observe that the elemental phosphorus exists as a tetratomic molecule.( tetra atomic molecules are those molecules which are formed by a combination of four atoms.)
Now, let us know about Gay-Lussac’s law, because we will be using this concept in our further steps.
- Gay-Lussac’s law is a gas law which states that the pressure exerted by a gas varies directly with the accurate temperature of a gas. In simple words the pressure exerted by a gas is proportional to the temperature of the gas when the mass is fixed and the volume is constant.
$P \propto T{\text{ and }}\frac{P}{T} = k$
Here,
- P is the pressure of the gas phosphene
- T is the absolute temperature of gas
- K is a constant.
From the reaction, we can say that, when the temperature and pressure remain the same, 4ml of $P{H_3}$ will give 6ml of ${H_2}$ (by applying Gay-Lussac’s law).
- We can say that 100ml of $P{H_3}$ will yield 150ml of ${H_2}$.
Therefore,
Increase in volume = (150-100)ml =50ml
Here, the volume of solid phosphorus is negligible).
Hence, the increase in volume is 50ml.
So, the correct answer is “Option A”.

Note: Some, real-life examples of Gay-Lussac’s law are the rupture of a pressure cooker, aerosol can and a tyre. All these substances explode when exposed to higher temperatures.