To solve the problem of finding the temperature at the bottom of the lake, we will use the Ideal Gas Law, which states that:
\[
\frac{P_A V_A}{T_A} = \frac{P_B V_B}{T_B}
\]
Where:
- \( P_A \) and \( P_B \) are the pressures at the bottom and surface of the lake, respectively.
- \( V_A \) and \( V_B \) are the volumes of the air bubble at the bottom and surface, respectively.
- \( T_A \) and \( T_B \) are the temperatures at the bottom and surface, respectively.
### Step 1: Calculate the pressure at the surface (P_B)
The pressure at the surface of the lake is equal to the atmospheric pressure, which is given as:
\[
P_B = 76 \, \text{cm of Hg}
\]
To convert this to dynes/cm² (CGS system), we use:
\[
1 \, \text{cm of Hg} = 13.6 \times 980 \, \text{dynes/cm}^2
\]
Thus,
\[
P_B = 76 \times 13.6 \times 980 \, \text{dynes/cm}^2
\]
### Step 2: Calculate the pressure at the bottom (P_A)
The pressure at the bottom of the lake includes both the atmospheric pressure and the hydrostatic pressure due to the water column above it. The hydrostatic pressure can be calculated using:
\[
P_A = P_B + \rho g h
\]
Where:
- \( \rho \) (density of water) = \( 1 \, \text{g/cm}^3 \)
- \( g \) = \( 980 \, \text{cm/s}^2 \)
- \( h \) = \( 250 \, \text{cm} \)
Thus,
\[
P_A = P_B + (1 \times 980 \times 250) \, \text{dynes/cm}^2
\]
### Step 3: Calculate the volumes (V_A and V_B)
The volume of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
- At the bottom (diameter = 3.6 mm = 0.36 cm), the radius \( r_A = \frac{3.6}{2} = 1.8 \, \text{mm} = 0.18 \, \text{cm} \):
\[
V_A = \frac{4}{3} \pi (0.18)^3
\]
- At the surface (diameter = 4 mm = 0.4 cm), the radius \( r_B = \frac{4}{2} = 2 \, \text{mm} = 0.2 \, \text{cm} \):
\[
V_B = \frac{4}{3} \pi (0.2)^3
\]
### Step 4: Convert the temperature at the surface (T_B) to Kelvin
Given:
\[
T_B = 40^\circ C = 40 + 273 = 313 \, K
\]
### Step 5: Substitute values into the Ideal Gas Law
Now we can substitute \( P_A, V_A, T_A, P_B, V_B, T_B \) into the Ideal Gas Law equation:
\[
\frac{P_A V_A}{T_A} = \frac{P_B V_B}{T_B}
\]
Rearranging for \( T_A \):
\[
T_A = \frac{P_A V_A T_B}{P_B V_B}
\]
### Step 6: Calculate \( T_A \)
Substituting the calculated values of \( P_A, V_A, P_B, V_B, T_B \) into the equation will yield \( T_A \).
### Step 7: Convert \( T_A \) back to Celsius
Finally, convert \( T_A \) from Kelvin back to Celsius:
\[
T_A = T_A - 273
\]
### Final Answer
After performing the calculations, we find:
\[
T_A \approx 10.37^\circ C
\]
