Question

A bubble of volume ${{V}_{1}}$ is at the bottom of a pond at $15{}^\circ C$ and 1.5 atm pressure. When it comes at the surface, it observes a pressure of 1 atm at $25{}^\circ C$ and has volume ${{V}_{2}}$ , given $\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)$ .
(A) 15.5
(B) 0.155
(C) 155.0
(D) 1.55

Answer 1

Hint: The general idea about the ideal gas equations and the values will help us solve the given illustration. As the final answer will be the ratio of the volumes, there will be no unit and also, any one of all the options can be correct.

Complete Solution :
Let us study the ideal gas equation before moving towards the given illustration:
Ideal gas equation-
An ideal gas is a hypothetical gas following all general laws such as Boyle’s law and Charles law which then forms the universally acceptable ideal gas equation i.e.
\[PV=nRT\]
where,
P = pressure of an ideal gas
V = volume of an ideal gas
n = the amount of ideal gas measured in terms of moles
R = universal gas constant
T = temperature
So, when we compare two equimolar gases at conditions where pressure, temperature and volumes are comparable:

We can ideally conclude that,
\[\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\]
Thus, now solving for our illustration:
We have,
${{P}_{1}}$ = 1.5 atm
${{P}_{2}}$ = 1 atm
${{T}_{1}}$ = 15 + 273 = 288 K
${{T}_{2}}$ = 25 + 273 = 298 K

Applying the given formula:
\[\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\]
$\therefore \dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{P}_{1}}{{T}_{2}}}{{{P}_{2}}{{T}_{1}}}$
Thus, $\dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{1.5\times 298}{1\times 288}=1.55$
So, the correct answer is “Option D”.

Note: Do note that the temperature value in the ideal gas equation will be in kelvins. So, before solving conversion is necessary.
- Also, as we have to find the ratio of similar quantities, thus there will be no unit for the final answer of the illustration.