an air bubble is released at the bottom of a lake where the temperature is 4 celsius and the pressure is 3.40 atm. if the bubble was 10.0 ml to start, what well its volume be at the surface, where the water temperature is 12 celsius and the pressure is 103 kpa?
(P1V1)/T1 = (P2V2)/T2
You need to change atm to kPa or the other way around.
123 kpa
To determine the volume of the air bubble at the surface of the lake, we can use the combined gas law, which is a rearranged version of the ideal gas law. The combined gas law relates the initial and final states of a gas sample under different conditions of temperature, pressure, and volume.
The combined gas law formula is:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
where:
P₁ and P₂ are the initial and final pressures, respectively,
V₁ and V₂ are the initial and final volumes, respectively,
T₁ and T₂ are the initial and final temperatures, respectively.
Now let's plug in the given values into the formula:
P₁ = 3.40 atm (the pressure at the bottom of the lake)
V₁ = 10.0 ml (the starting volume of the bubble)
T₁ = 4 °C + 273.15 = 277.15 K (convert Celsius to Kelvin)
P₂ = 103 kPa (the pressure at the surface of the lake)
V₂ = ? (the volume at the surface)
T₂ = 12 °C + 273.15 = 285.15 K (convert Celsius to Kelvin)
Now let's rearrange the formula to solve for V₂:
V₂ = (P₁ * V₁ * T₂) / (P₂ * T₁)
Plugging in the given values:
V₂ = (3.40 atm * 10.0 ml * 285.15 K) / (103 kPa * 277.15 K)
Note: Convert kPa to atm by dividing by 101.325 (1 atm = 101.325 kPa)
V₂ = (3.40 atm * 10.0 ml * 285.15 K) / (103 kPa * 277.15 K / 101.325)
V₂ = (3.40 atm * 10.0 ml * 285.15 K * 101.325) / (103 kPa * 277.15 K)
Calculating the above expression will yield the volume of the air bubble at the surface of the lake.