The final temperature is approximately \(-272.37 \, \text{°C}\).
To solve this problem, you can use the combined gas law, which is expressed as:
\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \]
Where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively,
- \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively,
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
First, convert the temperatures from Celsius to Kelvin using the relationship \( T(K) = T(°C) + 273.15 \).
Given initial conditions:
- \( V_1 = 74.12 \, \text{L} \)
- \( T_1 = 41.2 + 273.15 \, \text{K} \)
- \( P_1 = 764 \, \text{mm Hg} \)
And final conditions:
- \( V_2 = 44.27 \, \text{L} \)
- \( P_2 = 1002 \, \text{mm Hg} \)
Now plug in these values into the combined gas law:
\[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \]
Solve for \( T_2 \):
\[ T_2 = \frac{P_2 \cdot V_2 \cdot T_1}{P_1 \cdot V_1} \]
Substitute the values:
\[ T_2 = \frac{(1002 \, \text{mm Hg}) \cdot (44.27 \, \text{L}) \cdot (41.2 + 273.15 \, \text{K})}{(764 \, \text{mm Hg}) \cdot (74.12 \, \text{L})} \]
Now calculate:
\[ T_2 \approx \frac{44326.4}{56779.68} \, \text{K} \]
\[ T_2 \approx 0.78 \, \text{K} \]
Convert back to Celsius:
\[ T_2 \approx -272.37 \, \text{°C} \]
So, the final temperature is approximately \(-272.37 \, \text{°C}\).