A compound has a Normal Boiling Point of 95.0 °C and an Enthalpy of Vaporization of 42.0 kJ/mol. What is the Vapor Pressure (in atm) when the compound is at 105.0 °C? (R = 8.314 J/Kmol)
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
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![### Problem Statement:
A compound has a Normal Boiling Point of 95.0 °C and an Enthalpy of Vaporization of 42.0 kJ/mol. What is the Vapor Pressure (in atm) when the compound is at 105.0 °C? (R = 8.314 J/Kmol)
**Note:**
- The Normal Boiling Point is the temperature at which the vapor pressure of the liquid equals the atmospheric pressure (1 atm).
- Enthalpy of Vaporization is the energy required for the phase transition from liquid to gas.
- R represents the universal gas constant.
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**Calculation:**
To solve for the vapor pressure at 105.0 °C, you may use the Clausius-Clapeyron equation:
\[ \ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \]
Where:
- \( P_1 \) = Vapor pressure at the boiling point (1 atm for normal boiling point)
- \( P_2 \) = Vapor pressure at the desired temperature
- \( \Delta H_{vap} \) = Enthalpy of Vaporization
- \( R \) = Gas constant (8.314 J/K\u00b7mol)
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin
Convert temperatures from Celsius to Kelvin:
\[ T_1 = 95.0°C + 273.15 \]
\[ T_2 = 105.0°C + 273.15 \]
### Diagram:
This problem does not have any accompanying graphs or diagrams. However, it instructs the student to apply the Clausius–Clapeyron equation to solve for the vapor pressure at a given temperature.
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