
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
573k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

