Question

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep a temperature of 12° C. To what volume does it grow when it reaches the surface, which is at a temperature of 35° C?

Answer 1

use the formula P,V,T =constant.
v1= 1.0
v2=?
T1 =12+273K
T2=35+273
P1=h(rho)g
P2=atm presssure
hence u will get ur answer

Answer 2

Hey dear,


◆ Answer- 5.263 × 10^-6 m^3

◆ Explaination-
# Given-
V1 = 1 cm^3 = 10^–6 m^3
h = 40 m
T1 = 12 °C = 285 K
T2 = 35 °C = 308 K
P1 = ?
P2 = 1 atm = 1.013×10^5 Pa

# Solution-
Here pressure at depth of 40 m will be
P1 = P2 + ρhg
P1 = 1.013×10^5 + 1000×40×9.8
P1 = 10130 + 39200
P1 = 493300 Pa

We have the relationship,
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1P2
V2 = 493300 × 10^-6 × 308 / 285 × 10130
V2 = 5.263 × 10^-6 m^3


Therefore, volume of the air bubble will be 5.263×10^-6 m^3.

Hope this helps ...