Question

A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the bubble when it reaches the surface where the pressure is 1.0 atm and the temperature is 27 °C, if the moles are constant?

Answer 1

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 2.4 atm

$P_2$ = final pressure of gas = 1.0 atm

$V_1$ = initial volume of gas = 250 mL

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $15^oC=273+15=288K$

$T_2$ = final temperature of gas = $27^oC=273+27=300K$

Now put all the given values in the above equation, we get:

$\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}$

$V_2=625mL$

Therefore, the volume of the bubble is, 625 mL