Question

We have learned that the enthalpy of vaporization of a liquid is generally a function of temperature. If we wish to take this temperature variation into account, we cannot use the Clausius-Clapeyron equation in the form given in the text (that is, equation 12.2 ). Instead, we must go back to the differential equation upon which the Clausius-Clapeyron equation is based and reintegrate it into a new expression. Our starting point is the following equation describing the rate of change of vapor pressure with temperature in terms of the enthalpy of vaporization, the difference in molar volumes of the vapor \(\left(V_{g}\right),\) and liquid \(\left(V_{1}\right),\) and the temperature. $$\frac{d P}{d T}=\frac{\Delta H_{\mathrm{vap}}}{T\left(V_{\mathrm{g}}-V_{1}\right)}$$ Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the \(V_{1}\) term as if it were zero. Also, unless the vapor pressure is unusually high, we can treat the vapor as if it were an ideal gas; that is, for one mole of vapor, \(P V=R T\). Make appropriate substitutions into the above expression, and separate the \(P\) and \(d P\) terms from the \(T\) and \(d T\) terms. The appropriate substitution for \(\Delta H_{\text {vap }}\) means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits \(P_{1}\) and \(P_{2}\) on one side and \(T_{1}\) and \(T_{2}\) on the other. (a) Derive an equation for the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{l})\) as a function of temperature, if \(\Delta H_{\mathrm{vap}}=\) \(15,971+14.55 T-0.160 T^{2}\left(\text { in } J m o l^{-1}\right)\) (b) Use the equation derived in (a), together with the fact that the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(1)\) at \(120 \mathrm{K}\) is 10.16 Torr, to determine the normal boiling point of ethylene.

Short answer

The normal boiling point of ethylene, which is the temperature at which its vapor pressure equals 1 atm, is determined by solving the integrated version of the Clausius-Clapeyron equation with given enthalpy of vaporization, vapor pressure at 120K and R constant.

Step-by-step solution

Start with the given equation

Start with the given equation \( \frac{d P}{d T}=\frac{\Delta H_{\mathrm{vap}}}{T(V_{\mathrm{g}}-V_{1})} \). Assuming \(V_1\) is approximately zero and using the ideal gas law \(P V = RT\), we have \(V_g = \frac{RT}{P}\). Substitute these into the equation, which simplifies to \( \frac{d P}{d T}=\frac{P \Delta H_{vap}}{RT^2} \)

Separate variables

Now we separate the \(P\) and \(d P\) terms from the \(T\) and \(dT\) terms to create an expression suitable for integrating. This gives \( \frac{d P}{P}=\frac{\Delta H_{vap}}{RT^2} dT. \)

Substitute for ΔHvap and integrate

Substitute \(\Delta H_{vap}\) as a function of temperature, so the expression becomes \( \frac{d P}{P}=\frac{15971 + 14.55T - 0.16T^2}{RT^2} dT \). Integrate both sides between \(P_1\) and \(P_2\) and between \(T_1\) and \(T_2\) respectively.

Solve for P2 (the boiling point pressure)

After integrating and simplifying, you should have an equation in the form of \( ln(P_2/P_1) = \int_{T_1}^{T_2} \frac{15971 + 14.55t - 0.16t^2}{Rt^2} dt \). Given that the vapor pressure of ethylene at 120K is 10.16 Torr (or about 1.3417 kPa), we can set \(P_1\) to this value and \(T_1\) to 120. Solving for \(P_2\) when \(T_2\) is x will give the vapor pressure at any temperature x.

Finding the normal boiling point

The normal boiling point of a substance is the temperature at which its vapor pressure is equal to 1 atmosphere. So this can be calculated by solving \( ln(P_2/P_1) = \int_{120}^{T_2} \frac{15971 + 14.55t - 0.16t^2}{Rt^2} dt \) where \(P_2 = 1\) atm, for \(T_2\).

Key concepts

Enthalpy of Vaporization

The enthalpy of vaporization is a crucial concept when studying phase changes, particularly from liquid to vapor. It refers to the amount of energy required to transform one mole of a liquid into its gaseous state at constant pressure.
This energy overcomes the intermolecular forces between liquid particles to allow them to enter the vapor phase.
Enthalpy of vaporization is often expressed as: \[\Delta H_{\text{vap}} = H_{\text{vapor}} - H_{\text{liquid}}\]Where \(\Delta H_{\text{vap}}\) is the enthalpy change, \(H_{\text{vapor}}\) is the enthalpy of the gaseous phase, and \(H_{\text{liquid}}\) is the enthalpy of the liquid phase. This value is usually temperature-dependent due to changes in intermolecular forces and system entropy with temperature.
Understanding the enthalpy of vaporization helps in explaining why different liquids have different boiling points and vapor pressures, which are essential variables in processes like distillation or in climate modeling for determining water vapor content in the atmosphere.

Vapor Pressure

Vapor pressure is a term used to describe the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. It signifies the tendency of particles to escape from the liquid to the vapor state.
When the vapor pressure equals the atmospheric pressure, the liquid begins to boil.
Key features of vapor pressure include:

• Temperature Dependency: Vapor pressure rises with an increase in temperature since higher temperatures give particles more energy to escape the liquid phase.
• Intermolecular Forces: Weak intermolecular forces result in higher vapor pressures, as molecules can more easily leave the liquid.
• Clausius-Clapeyron Relationship: This equation relates vapor pressure and temperature. It is an exponential relationship indicating how vapor pressure changes with enthalpy of vaporization and temperature.

Understanding vapor pressure is important in a variety of applications, such as meteorology and engineering, where controlling vapor-liquid equilibria is crucial.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics, relating the four state variables: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and number of moles (\(n\)) by the equation:
\[ PV = nRT \]Here, \(R\) is the universal gas constant. This equation models the behavior of "ideal" gases, which are hypothetical gases that perfectly obey the assumptions of the kinetic molecular theory under a wide range of conditions.
Key aspects of the Ideal Gas Law:

• Simplified Model: While real gases may deviate from ideal behavior at high pressures and low temperatures, this equation provides a good approximation under standard conditions.
• Applications: Used for calculating changes in a gas system, designing equipment like engines, and understanding atmospheric phenomena.

Understanding the Ideal Gas Law helps in explaining properties of gases and their interactions in different conditions. It's integral in applications ranging from calculating the molar volume of a gas to scuba diving gas computations.