Question

An air bubble rises from the bottom of a well where the temperature is 25°C to the surface where the temperature is 27°C. Find the percent increase in the volume of the bubble if the depth of the well is 5 m. Atmospheric pressure is 101,528 Pascals.    

Answer 1

We have to consider the ideal gas law, thus the relation between both states will be given by

$\cfrac{P_1V_1}{T_1} = \cfrac{P_2V_2}{T_2} \implies \cfrac{V_2}{V_1} = \cfrac{P_1 T_2}{P_2T_1}$

Now, we need to define the pressures and temperatures to calculate the rate between the new volume out of the well (V₂) and the original volume of gas at the bottom (V₁):

$P_1=P_{bottom}=\rho_{_{\text{H}_2\text{O}}}gh \\ P_1= 1000\frac{kg}{m^3}\cdot 9.80\frac{m}{s^2}\cdot 5\,m\times \frac{1\,Pa}{1\,kg/(m^2 \cdot s^2)} \\P_1=49000\,Pa \\ T_1=T_{bottom}=25\,°C=298.15\,K \\ P_2=P_{surface}=P_{atm} \\ P_2=101528\,Pa \\ T_2=T_{surface}=27\,°C=300.15\,K$

Now we substitute and we will find the change in volume:

$\text{\% of change} = 100\cdot \cfrac{V_2-V_1}{V_1} = 100\cdot \Bigg( \cfrac{P_1 T_2}{P_2T_1} -1\Bigg) \\ \text{ } \\ \text{\% of change} = 100\cdot \Bigg( \frac{(101528\text{ Pa}) (300.15\text{ K})}{(49000\text{ Pa}) (298.15\text{ K})} -1\Bigg) \\ \therefore \text{\% of change} = 100\cdot (2.0589-1)=105.89\text{ \%}$

In conclusion, the volume of the bubble on the surface will increase and be 105.89 % higher than the volume at the bottom V₁.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.