An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?
Given: The volume of air bubble is 1.0 cm 3 , the depth from which air bubble rises is 40 m , temperature of bubble inside the lack is 12 ° C and on the surface of lack is 35 ° C .
The pressure at a certain depth is given as,
P 1 = 1 atm + ρ g h
Where, ρ is the density of water, g is the acceleration due to gravity and h is the depth.
By substituting the values in the above equation, we get
P 1 = 1.013 × 10 5 + 1000 × 9.8 × 40 = 4.933 × 10 5 Pa
The combined gas law is given as,
P 1 V 1 T 1 = P 2 V 2 T 2 V 2 = P 1 V 1 T 1 × T 2 P 2
Where, P 1 , V 1 and T 1 are the pressure, volume and temperature inside the lack and P 2 , V 2 and T 2 are the pressure, volume and temperature at the surface.
By substituting the given values in the above equation, we get
V 2 = 4.933 × 10 5 ( 1.0 × 10 − 6 ) × ( 35 + 273 ) ( 12 + 273 ) × ( 1.013 × 10 5 ) = 5.26 × 10 − 6 m 3
Thus, the volume at surface is 5.26 × 10 − 6 m 3 .
Given: The volume of air bubble is
The pressure at a certain depth is given as,
Where,
By substituting the values in the above equation, we get
The combined gas law is given as,
Where,
By substituting the given values in the above equation, we get
Thus, the volume at surface is
