Answer: To solve this problem, we can use the ideal gas law equation:
\[ P_1/T_1 = P_2/T_2 \]
where:
- \( P_1 \) and \( T_1 \) are the initial pressure and temperature, respectively.
- \( P_2 \) and \( T_2 \) are the final pressure and temperature, respectively.
Given:
- \( P_1 = 10 \) mm Hg (at the triple point of water).
- \( T_1 = 0 \) degrees Celsius (at the triple point of water).
- \( T_2 = 50 \) degrees Celsius.
First, we need to convert the temperatures to Kelvin since the ideal gas law requires temperatures in Kelvin.
\[ T_1 = 0 + 273.15 = 273.15 \text{ K} \]
\[ T_2 = 50 + 273.15 = 323.15 \text{ K} \]
Now, let's solve for \( P_2 \):
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]
\[ P_2 = 10 \times \frac{323.15}{273.15} \]
\[ P_2 = 10 \times 1.1837 \]
\[ P_2 \approx 11.837 \text{ mm Hg} \]
Therefore, the pressure at 50 degrees Celsius will be approximately 11.837 mm Hg.
