Question

A bubble of air is underwater at temperature ${{15}^{o}}C$ and the pressure $1.5$ bar. If the bubble rises to the surface where the temperature is ${{25}^{o}}C$ and the pressure is $1.0$ bar, what will happen to the volume of the bubble?
A. Volume will become greater by a factor of $1.6$
B. Volume will become greater by a factor of $1.1$
C. Volume will become smaller by a factor of $0.70$
D. Volume will become greater by a factor of $2.5$

Answer 1

Hint: According to the question, the initial and final temperature and initial and final pressure are already, we have to find out the change in volume. To get a suitable answer, consider the ideal gas equation, where pressure, volume and temperature are related.

Complete step by step answer:
As per the given question, a bubble of air is underwater at temperature ${{15}^{o}}C$$1.0$and the pressure equals $1.5$ bar. Then the bubble rises to the surface where its temperature equals ${{25}^{o}}C$ and the pressure is $1.0$ bar. And, we have to find out the change in volume of the bubble.

So, let us consider;
The initial temperature is ${{T}_{1}}$, which is equal to ${{15}^{o}}C$ i.e. $(15+273)K=288K$.
The final temperature is ${{T}_{2}}$, which is given as ${{25}^{o}}C$ i.e. $(25+273)K=298K$.
The initial pressure is ${{P}_{1}}$, whose value is given as $1.5$bar.
The final pressure is ${{P}_{2}}$, whose value is $1.0$ bar.
The initial volume be ${{V}_{1}}$ and the final volume be ${{V}_{2}}$.
The ideal gas equation states that,
$PV=nRT$, where pressure (P) and volume (V) are directly proportional to the temperature (T).
The ideal gas equation for two different condition can be represented as,
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$
So, now by substituting the given values in the above formula, we get:
$\dfrac{1.5\times {{V}_{1}}}{288}=\dfrac{1.0\times {{V}_{2}}}{298}$
Then, $\dfrac{1.5\times 298}{1.0\times 288}=\dfrac{{{V}_{2}}}{{{V}_{1}}}$
So, $\dfrac{{{V}_{2}}}{{{V}_{1}}}=1.55$
And, ${{V}_{2}}=1.55{{V}_{1}}\cong 1.6{{V}_{1}}$
Therefore, the final volume will be $1.6$ times the initial volume i.e. the volume of the bubble will become greater by a factor of $1.6$.
So, the correct answer is “Option A”.

Note: An ideal gas is a gas that obeys Boyle’s law, Charle’s law and Avogadro’s law at variable pressure, volume and temperature. And this law is referred to as the ideal gas law. The ideal gas law tends to fall at lower temperature and higher pressures and also fails for most of the heavy gases.