Question

13. A bubble of air is underwater at temperature 15°Cand the pressure 1.5 bar.
If the bubble rises to thesurface where the temperature is 25°C and thepressure is 1.0 bar what will happen to the volume ofthe bubble ?

[AIPMT (Mains)-2011]

(1) Volume will become smaller by a factor of 0.70

(2) Volume will become greater by a factor of 2.5

(3) Volume will become greater by a factor of 1.6

(4) Volume will become greater by a factor of 1.1​

Answer 1

Answer : The correct option is, (3) Volume will become greater by a factor of 1.6

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 = 1.5 bar

$P_2$ = final pressure of gas = 1.0 bar

$V_1$ = initial volume of gas = V

$V_2$ = final volume of gas = $V_2$

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

$T_2$ = final temperature of gas = $25^oC=273+25=298K$

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

$\frac{1.5bar\times V}{288K}=\frac{1.0bar\times V_2}{298K}$

$V_2=1.55V\approx 1.6V$

Therefore, the volume will become greater by a factor of 1.6