
A bubble of air is underwater at temperature ${15^\circ }C$ and pressure $1.5bar$. If the bubble rises to the surface where the temperature is ${25^\circ }C$ and the pressure is $1.0bar$ , 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
550.2k+ views
Hint: Focus on the given parameters, where you might find the gas equation that suits fit for determining volume changes. Initial and final conditions need to be separately differentiated.
Complete step by step answer:
The initial condition is given with the pressure $P_1$ which is defined here as $1.5bar$ for the bubble when it is under water. The initial volume can be considered as $V_1$ for the bubble and the temperature underwater which can be considered as initial temperature $T_1$ is about ${15^\circ }C$ or $288K$.
The final condition of the bubble when it comes towards the surface, the temperature changes to become the final temperature $T_2$ which is ${25^\circ }C$ or $298K$ and the pressure $P_2$ changes to $1.0bar$.
Hence two specific states are there in which the bubbles are present with separate physical properties in those states. This is the importance of gas equation and
Therefore, according to the ideal gas equation:
$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Putting all the given values in this equation it can be determined that
$\dfrac{{1.5 \times {V_1}}}{{288}} = \dfrac{{1 \times {V_2}}}{{298}}$
${V_2} = 1.55{V_1}$
Therefore, the volume change that occurs after the bubble moves up to the surface of the water is about $1.6$ times that of the initial volume. Hence the change in volume as observed based on the changing physical conditions from the internal underwater condition to the surface of the water. The bubble rises up and the volume of the bubble is expected to increase with it by the factor of $1.6$ times.
So, the correct answer is Option A.
Note: While determining the changes from the ideal gas equation, it is most important to convert the temperature in Kelvin scale as the equation follows that specific scale only. The other important thing is while comparing, finding the factor of one for the other is the best possible option.
Complete step by step answer:
The initial condition is given with the pressure $P_1$ which is defined here as $1.5bar$ for the bubble when it is under water. The initial volume can be considered as $V_1$ for the bubble and the temperature underwater which can be considered as initial temperature $T_1$ is about ${15^\circ }C$ or $288K$.
The final condition of the bubble when it comes towards the surface, the temperature changes to become the final temperature $T_2$ which is ${25^\circ }C$ or $298K$ and the pressure $P_2$ changes to $1.0bar$.
Hence two specific states are there in which the bubbles are present with separate physical properties in those states. This is the importance of gas equation and
Therefore, according to the ideal gas equation:
$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Putting all the given values in this equation it can be determined that
$\dfrac{{1.5 \times {V_1}}}{{288}} = \dfrac{{1 \times {V_2}}}{{298}}$
${V_2} = 1.55{V_1}$
Therefore, the volume change that occurs after the bubble moves up to the surface of the water is about $1.6$ times that of the initial volume. Hence the change in volume as observed based on the changing physical conditions from the internal underwater condition to the surface of the water. The bubble rises up and the volume of the bubble is expected to increase with it by the factor of $1.6$ times.
So, the correct answer is Option A.
Note: While determining the changes from the ideal gas equation, it is most important to convert the temperature in Kelvin scale as the equation follows that specific scale only. The other important thing is while comparing, finding the factor of one for the other is the best possible option.
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