To determine the temperature \( T_2 \) at which water will boil under the pressure of 5 times its vapor pressure, we need to use the Clausius-Clapeyron equation. Let's break down the steps to find the boiling temperature \( T_2 \):

### Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is:

\[
\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\]

Where:
- \( P_2 \) = 5 times the vapor pressure at \( T_2 \)
- \( P_1 \) = Vapor pressure at \( T_1 \) (usually 100°C or another known temperature)
- \( \Delta H_{\text{vap}} \) = Heat of vaporization of water (approximately \( 40.7 \, \text{kJ/mol} \))
- \( R \) = Gas constant \( 8.314 \, \text{J/mol·K} \)
- \( T_1 \) = Known temperature (in Kelvin)
- \( T_2 \) = Temperature at which the vapor pressure is \( P_2 \)

### Rearranging the Equation

We can rearrange the equation to solve for \( T_2 \):

\[
\frac{1}{T_2} = \frac{1}{T_1} + \frac{R}{\Delta H_{\text{vap}}} \ln\left(\frac{P_1}{P_2}\right)
\]

Since \( P_2 = 5P_1 \):

\[
\frac{1}{T_2} = \frac{1}{T_1} + \frac{R}{\Delta H_{\text{vap}}} \ln\left(\frac{1}{5}\right)
\]

\[
\frac{1}{T_2} = \frac{1}{T_1} - \frac{R \ln(5)}{\Delta H_{\text{vap}}}
\]

\[
T_2 = \frac{1}{\frac{1}{T_1} - \frac{R \ln(5)}{\Delta H_{\text{vap}}}}
\]

### Numerical Calculation

Let's assume \( T_1 = 373.15 \, \text{K} \) (which is 100°C or 212°F) where the vapor pressure is 1 atm:

\[
T_2 = \frac{1}{\frac{1}{373.15} - \frac{8.314 \times \ln(5)}{40700}}
\]

Now, calculate the values:

1. \( \ln(5) \approx 1.609 \)
2. \( \frac{R \ln(5)}{\Delta H_{\text{vap}}} = \frac{8.314 \times 1.609}{40700} \approx 0.000329 \)
3. \( \frac{1}{T_1} = \frac{1}{373.15} \approx 0.00268 \)
4. \( \frac{1}{T_2} = 0.00268 - 0.000329 \approx 0.002351 \)

Finally:

\[
T_2 \approx \frac{1}{0.002351} \approx 425.32 \, \text{K}
\]

### Final Answer

The temperature \( T_2 \) at which water boils under 5 times the vapor pressure is approximately **425.32 K** (or about 152.17°C).

Would you like more details on the calculations or the implications of these results?

Here are some related questions to consider:

1. How does the boiling point change with different pressures?
2. What is the significance of the heat of vaporization?
3. How does this temperature compare to typical boiling points at different altitudes?
4. What would happen if the pressure were lower instead of higher?
5. How does this relationship apply to industrial applications?

**Tip:** Remember that boiling point elevation and depression can be used to control processes in both industrial and laboratory settings!

Learn how to calculate the boiling temperature \( T_2 \) of water under 5 times its vapor pressure using the Clausius-Clapeyron equation. Includes step-by-step calculation details and implications. Temperature calculated to approximately 425.32 K (152.17°C).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation