Question

A 0.5-cm diameter air bubble at 5°C and 3 atm pressure rises to the surface. Determine its final diame- ter when T 25°C and P 1 atm. (Ignore any water vapor in the bubble.)

Answer 1

Answer:

d=0.738cm

Explanation:

We divide this problem in two parts, below the surface is part 1 (where we have $P_1V_1=n_1RT_1$), and on the surface part 2 (where we have $P_2V_2=n_2RT_2$).  In this case, pressure, volume and temperature will vary, and n will remain constant because the amount of air is always the same (and R is always a constant), so we have $n_1=n_2$.

We can then do:

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

The volume of a sphere is $V=\frac{4\pi r^3}{3}$, so we can write

$\frac{P_1}{T_1}\frac{4\pi r_1^3}{3}=\frac{P_2}{T_2}\frac{4\pi r_2^3}{3}$

Which is

$\frac{P_1r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}$

And obtain our radius for the bubble on the surface:

$r_2=\sqrt[3]{\frac{P_1T_2r_1^3}{P_2T_1}}=\sqrt[3]{\frac{P_1T_2}{P_2T_1}}r_1$

Since the diameter is 2 times the radius this can be written as (the factor 2 cancels out):

$d_2=\sqrt[3]{\frac{P_1T_2}{P_2T_1}}d_1$

And we put our values in S.I.:

$d_2=\sqrt[3]{\frac{(3atm)(298K)}{(1atm)(278K)}}(0.005m)=0.00738m=0.738cm$