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A certain substance has a heat of vaporization of 39.96 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 305 K?

A certain substance has a heat of vaporization of 39.96 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 305 K?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:** A certain substance has a heat of vaporization of 39.96 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 305 K? **Box for Answer:** \[ T = \quad \text{K} \] **Explanation:** To solve this problem, you would typically use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature. The 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: - \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively. - \( \Delta H_{\text{vap}} \) is the heat of vaporization (39.96 kJ/mol). - \( R \) is the ideal gas constant (8.314 J/mol·K). - \( T_1 \) is the initial temperature (305 K). - \( T_2 \) is the final temperature you need to find. Since \( P_2 = 7.00 \times P_1 \), substitute into the equation and solve for \( T_2 \).
Transcribed Image Text:**Problem Statement:** A certain substance has a heat of vaporization of 39.96 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 305 K? **Box for Answer:** \[ T = \quad \text{K} \] **Explanation:** To solve this problem, you would typically use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature. The 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: - \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively. - \( \Delta H_{\text{vap}} \) is the heat of vaporization (39.96 kJ/mol). - \( R \) is the ideal gas constant (8.314 J/mol·K). - \( T_1 \) is the initial temperature (305 K). - \( T_2 \) is the final temperature you need to find. Since \( P_2 = 7.00 \times P_1 \), substitute into the equation and solve for \( T_2 \).
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