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If the vapor pressure of ethy1 alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is \(0.132 \mathrm{~atm}\) at \(34.9^{\circ} \mathrm{C}\), and \(0.526 \mathrm{~atm}\) at \(63.5^{\circ} \mathrm{C}\) what do you predict it will be at \(19.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The predicted vapor pressure of ethyl alcohol at 19.0°C is approximately 0.069 atm.

Step by step solution

01

Write down the Clausius-Clapeyron equation

The Clausius-Clapeyron equation is \[ \ln\frac{P_2}{P_1} = \frac{-\Delta H_v}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), respectively, \(\Delta H_v\) is the heat of vaporization, and R is the universal gas constant (8.314 J/mol·K).
02

Solve the Clausius-Clapeyron equation for the unknown vapor pressure

Let's denote the vapor pressures at 34.9°C and 63.5°C as \(P_1\) and \(P_2\), respectively, and the vapor pressure at 19.0°C as \(P\). We will solve the Clausius-Clapeyron equation for \(P\), using the known values of \(P_1\), \(P_2\), \(T_1\), and \(T_2\). To do this, we need to eliminate \(\Delta H_v\) from the equation. We can rewrite the Clausius-Clapeyron equation as \[ \Delta H_v = -R \frac{\ln(P_2/P_1)}{\frac{1}{T_2} - \frac{1}{T_1}} \] Now, we can substitute this expression for \(\Delta H_v\) into the Clausius-Clapeyron equation relating \(P\), \(P_1\) and their corresponding temperatures, to get \[ \ln\frac{P}{P_1} = \frac{-\Delta H_v}{R}\left(\frac{1}{T} - \frac{1}{T_1}\right) \] Substitute the expression for \(\Delta H_v\): \[ \ln\frac{P}{P_1} = \frac{\ln(P_2/P_1)}{\frac{1}{T_2} - \frac{1}{T_1}}\left(\frac{1}{T} - \frac{1}{T_1}\right) \]
03

Calculate the required values using the provided data

Let's convert the given temperatures from Celsius to Kelvin: \[ T_1 = 34.9 + 273.15 = 308.05\,\mathrm{K} \] \[ T_2 = 63.5 + 273.15 = 336.65\,\mathrm{K} \] \[ T = 19.0 + 273.15 = 292.15\,\mathrm{K} \] Now, we can plug in the given values for \(P_1 = 0.132\,\mathrm{atm}\) and \(P_2 = 0.526\,\mathrm{atm}\) and the calculated values for \(T_1\), \(T_2\), and \(T\) into the equation for \(P/P_1\): \[ \ln\frac{P}{0.132} = \frac{\ln(0.526/0.132)}{\frac{1}{336.65} - \frac{1}{308.05}}\left(\frac{1}{292.15} - \frac{1}{308.05}\right) \]
04

Predict the vapor pressure at the given temperature

Finally, we can solve the equation for \(P\): \[ P = 0.132 \times \exp \left[ \frac{\ln(0.526/0.132)}{\frac{1}{336.65} - \frac{1}{308.05}}\left(\frac{1}{292.15} - \frac{1}{308.05}\right) \right] \] \[ P \approx 0.069\,\mathrm{atm}\] The predicted vapor pressure of ethyl alcohol at 19.0°C is approximately 0.069 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vapor Pressure
Vapor pressure is a measure of the tendency of a substance to evaporate. It is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The vapor pressure of a substance increases with temperature, as more molecules have sufficient kinetic energy to escape from the liquid or solid phase into the vapor phase.

For example, when a liquid is heated, its molecules move faster, and some of them can overcome the intermolecular forces that hold them in the liquid phase. Those molecules that escape create a pressure above the liquid, which we refer to as vapor pressure. Understanding vapor pressure is crucial in many fields, including meteorology, engineering, and physical chemistry, because it affects processes like evaporation, boiling, and sublimation.
Heat of Vaporization
The heat of vaporization, denoted as \(\Delta H_v\), is the amount of heat energy required to convert a quantity of liquid into vapor without a change in temperature. It essentially measures the strength of the intermolecular forces; the higher the heat of vaporization, the stronger the forces keeping the molecules together in the liquid state. \(\Delta H_v\) is a critical component in various thermodynamic equations, including the Clausius-Clapeyron equation, which relates the vapor pressure and temperature of a substance.In practical terms, when you boil water, for instance, you're supplying heat to overcome the heat of vaporization so that the water molecules can escape into the air as steam. The heat of vaporization plays a pivotal role not only in phase changes but also in industry, where it is a factor in the efficiency of power plants and refrigeration systems.
Celsius to Kelvin Conversion
Temperature conversion between Celsius and Kelvin is a basic yet extremely important step in scientific calculations, especially in chemistry and physics where temperature plays a vital role. The Kelvin scale is an absolute thermodynamic temperature scale widely used in the scientific world. The key aspect to remember is that 0 K, or absolute zero, is the theoretical point where particles have minimal thermal motion.To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This stems from the definition that the difference in temperature by one degree Celsius is equivalent to the difference of one Kelvin. Thus, for example, a temperature of \( 25^\circ C \) is equivalent to \(25 + 273.15 = 298.15 K\) in Kelvin. Implementing this conversion correctly is crucial for the accuracy of scientific experiments and calculations, such as those using the Clausius-Clapeyron equation.
Natural Logarithm in Chemistry
The natural logarithm, commonly denoted as \(\ln\), is a mathematical function that's particularly useful in the science of chemistry. The \(\ln\) of a number is the power to which the base of the natural logarithms (which is \(e\), approximately equal to 2.71828) must be raised to obtain that number. In chemistry, natural logarithms are frequently employed in equations that describe the kinetics of chemical reactions, the distribution of molecules across different states of matter, and the temperature dependence on the rate of reactions, such as in the Arrhenius equation and the Clausius-Clapeyron equation.Understanding the proper use of the natural logarithm in chemistry is critical when interpreting reaction rates and the relationship between various thermodynamic parameters. It is also fundamental in the analysis of exponential decay and growth processes, enabling chemists to describe complex phenomena in a more manageable form.

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Most popular questions from this chapter

The freezing point constant of toluene is \(3.33^{\circ} \mathrm{C}\) per mole per \(1000 \mathrm{~g}\). Calculate the freezing point of a solution prepared by dissolving \(0.4\) mole of solute in \(500 \mathrm{~g}\) of toluene. The freezing point of toluene is \(-95.0^{\circ} \mathrm{C}\).

A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a \(50-50\) solution by volume of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\), ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is \(1.115\) and the freezing point depression constant of water is \(\left(1.86 \mathrm{C}^{\circ} / \mathrm{mole}\right)\). You might have placed in methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) instead of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\). If the current cost of ethylene glycol is 12 cents \(/ \mathrm{lb}\) and the cost of methanol is 8 cents \(/ \mathrm{lb}\), how much money would you save by using \(\mathrm{CH}_{3} \mathrm{OH}\) ? And yet, ethylene glycol is the more desirable antifreeze. Why? density \(=(.79 \mathrm{~g} / \mathrm{ml})\) for \(\mathrm{CH}_{3} \mathrm{OH}\) and \(3.785\) liter \(=1\) gallon.

If the vapor pressure of \(\mathrm{CC} 1_{4}\) (carbon tetrachloride) is \(.132\) atm at \(23^{\circ} \mathrm{C}\) and \(.526 \mathrm{~atm}\) at \(58^{\circ} \mathrm{C}\), what is the \(\Delta \mathrm{H}^{\prime}\) in this temperature range?

If the vapor pressure of methy1 alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), is \(0.0526\) atm at \(5.0^{\circ} \mathrm{C}\) and \(0.132 \mathrm{~atm}\) at \(21.2^{\circ} \mathrm{C}\), what do you predict the normal boiling point will be?

Assuming that the density of water is \(.9971\left(\mathrm{~g} / \mathrm{cm}^{3}\right)\) at \(25^{\circ} \mathrm{C}\) and that of ice at \(0^{\circ}\) is \(917\left(\mathrm{~g} / \mathrm{cm}^{3}\right)\), what percent of a water jug at \(25^{\circ} \mathrm{C}\) should be left empty so that, if the water freezes, it will just fill the jug?

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