Given:
The air pressure at sea level is
[tex]P_1=1.01\times10^5\text{ Pa}[/tex]The temperature at sea level is
[tex]\begin{gathered} T_1=17+273 \\ =290\text{ K} \end{gathered}[/tex]The air pressure at the final height is
[tex]P_2=0.85P_1[/tex]The final temperature is
[tex]\begin{gathered} T_2=6+273 \\ =279\text{ K} \end{gathered}[/tex]The radius of the bubble at the final height is
[tex]\begin{gathered} r_2=10\text{ cm} \\ =0.10\text{ m} \end{gathered}[/tex]To find:
The initial radius of the bubble
Explanation:
We know, for an ideal gas
[tex]\frac{PV}{T}=constant[/tex]We can write,
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \frac{P_1\times\frac{4}{3}\pi r_1^3}{T_1}=\frac{0.85P_1\frac{4}{3}\pi(r_2)^3}{T_2} \\ r_1^3=\frac{0.85\times(r_2)^3\times T_1}{T_2} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} r_1^3=\frac{0.85\times0.10^3\times290}{279} \\ r_1=\text{ 0.096 m} \\ r_1=9.6\text{ cm} \end{gathered}[/tex]Hence, the radius at sea level was 9.6 cm.
