a driver releases an air bubble of volume 2.0 cm^3 from a depth of 15 m below the surface of a lake where the temperature is 70C.

And then what? You have not asked a question. You will need to make an assumption about whether the temperature of the gas in the bubble stays the same.

To find the volume of the air bubble at the surface of the lake, we can use Boyle's Law, which states that the product of pressure and volume in a gas is constant if the temperature remains constant.

Boyle's Law equation:
P₁V₁ = P₂V₂

Where:
P₁ = initial pressure (pressure at depth)
V₁ = initial volume
P₂ = final pressure (pressure at the surface)
V₂ = final volume (volume at the surface)

Given:
Initial volume (V₁) = 2.0 cm³
Depth (h) = 15 m
Temperature (T) = 70°C

First, we need to calculate the initial pressure (P₁) using the hydrostatic pressure formula:

P₁ = ρgh

Where:
ρ = density of the liquid (water) = 1000 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = depth

Let's calculate P₁:
P₁ = (1000 kg/m³) * (9.8 m/s²) * (15 m)
P₁ = 147,000 Pa

Next, we need to convert the final temperature (T) from Celsius to Kelvin since Boyle's Law requires the temperature to be in Kelvin.

T(K) = T(°C) + 273.15
T(K) = 70°C + 273.15
T(K) = 343.15 K

Now, we can rearrange the Boyle's Law equation to find the final volume (V₂):

P₁V₁ = P₂V₂

V₂ = (P₁V₁) / P₂

For the final pressure (P₂), we can consider it as atmospheric pressure at the surface, usually around 101,325 Pa.

P₂ = 101,325 Pa

Let's substitute the values and calculate V₂:
V₂ = (147,000 Pa * 2.0 cm³) / 101,325 Pa
V₂ = 2.9 cm³

Therefore, the volume of the air bubble at the surface of the lake is approximately 2.9 cm³.