Question

A gas of volume 100cc is kept in a vessel at a pressure of ${{10}^{4}}$ Pa maintained at a temperature of ${{24}^{0}}C$. If the pressure is increased to ${{10}^{5}}$Pa keeping the temperature constant, then the volume (in cc) of gas will be:
(A) 1
(B) 10
(C) 100
(D) 1000

Answer 1

Hint:. The close approach for an answer to this question includes the formula for calculating volume of a gas based on the Boyle’s law which is given by the relationship ${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$ .

Complete step by step answer:
In our basic concepts of chemistry learnt in the lower classes, we have studied about the kinetic theory of gases and various laws relating to the gases.
- Boyle observed that the product of pressure and volume were found to be nearly constant for an ideal gas given by P$\times $ V= constant.

- Mathematically Boyle’s law can be written as pressure is inversely proportional to volume which is given as follows,
$P\propto \dfrac{1}{V}$

- Now, to calculate the above particular question, let us relate the pressure and volume of gas at constant temperature but changes in pressure based on Boyle’s law as,
${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$ .....(1)
From the data given, we have
${{V}_{1}} = 100cc$
\[{{P}_{1}} = {{10}^{4}}Pa\]
${{P}_{2}} = {{10}^{5}}Pa$ and the quantity to be found is ${{V}_{2}}$
Therefore, from equation number (1) we can rearrange it as,
${{V}_{2}} = \dfrac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}}$

By substituting the values from the data given we have,
${{V}_{2}} = \dfrac{{{10}^{4}}\times 100}{{{10}^{5}}}$
${{V}_{2}} = 10cc$
Thus, the volume of the gas when pressure is increased ${{10}^{5}}$Pa maintained at constant temperature will be 10 cc.
So, the correct answer is “Option A”.

Note: While solving such questions observe which quantity remains constant and not confuse with Charle’s law as it gives the relationship between volume and temperature when pressure is held constant whereas Boyle’s law gives the relationship between pressure and volume with temperature held constant.