Question

An underwater bubble with a radius of $0.5$cm at the bottom of the tank, where the temperature is ${\text{5}}{{\text{ }}^{\text{0}}}{\text{C}}$ and pressure is 3 atm rises to the surface, where the temperature ${\text{25}}{{\text{ }}^{\text{0}}}{\text{C}}$ and the pressure is 1 atm. What will be the radius of the bubble when it reaches the surface? Write numerical values upto 2 digits only after the decimal point.

Answer 1

This problem is based on the combined gas law equation that is based on three laws called the Boyle’s Law, the Charles’ Law, and the Avogadro’s Law and is applicable only to ideal gases. We shall use the pressure, volume and temperature dependence equation to calculate the volume and thus, the radius of the bubble.
Formula Used: $\dfrac{{{\text{PV}}}}{{\text{T}}} = {\text{constant}}$

Complete step by step solution
Given that, The temperature, ${{\text{T}}_{\text{1}}}$ of the bottom of the tank is ${\text{5}}{{\text{ }}^{\text{0}}}{\text{C}}$ and the pressure, ${{\text{P}}_{\text{1}}}$ is 3 atm. The initial radius of the bubble, ${{\text{r}}_{\text{1}}}$=$0.5$cm
The changed temperature, ${{\text{T}}_2}$ of the bottom of the tank is ${\text{25}}{{\text{ }}^{\text{0}}}{\text{C}}$ and the pressure, ${{\text{P}}_2}$ is 1 atm. The initial radius of the bubble, ${{\text{r}}_2}$=x cm.
Let the initial volume of the bubble = $\dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}_{\text{1}}}^{\text{3}}$ and the final volume = $\dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}_2}^{\text{3}}$
According to the combined gas law equation,
$\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$,
putting the values in their respective positions we get,
$\dfrac{{{{3 \times }}\dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\left( {0.5} \right)}^{\text{3}}}}}{{{\text{278}}}}{\text{ = }}\dfrac{{{{1}} \times \dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{x}}^{\text{3}}}}}{{298}}$
Rearranging:
Or, ${{\text{x}}^{\text{3}}} = \dfrac{{{\text{278}}}}{{3 \times 298 \times {{\left( {0.5} \right)}^{\text{3}}}}}$
Or, ${\text{x = 0}}{\text{.74}}$cm.
Hence, the radius of the bubble when it reaches the surface will be ${\text{0}}{\text{.74}}$cm.

Notes
An ideal gas is defined as one in which it is assumed that the volume of the gas molecules with respect to the volume of the gas is negligible and that there is no force of attraction or repulsion between the gas molecules. An ideal gas obeys the Boyle’s law that connects the pressure of the gas with the volume at constant temperature, the Charles’ Law that connects the volume of the gas with the Kelvin temperature scale at constant pressure and the Avogadro’s Law states that equal volumes of all gases contain equal number of moles at similar conditions of temperature and pressure.