Question

The normal boiling point of water is \(373 \mathrm{~K}\). Vapour pressure of water at temperature \(T\) is 19 \(\mathrm{mm} \mathrm{Hg}\). If enthalpy of vaporisation is \(40.67 \mathrm{~kJ} / \mathrm{mol}\), then temperature \(T\) would be (Use : \(\log 2=0.3, R: 8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ): (a) \(250 \mathrm{~K}\) (b) \(291.4 \mathrm{~K}\) (c) \(230 \mathrm{~K}\) (d) \(290 \mathrm{~K}\)

Short answer

The temperature T is approximately 291.4 K, so the correct option is (b) 291.4 K.

Step-by-step solution

Understand the Clausius-Clapeyron Equation

First, it's necessary to understand that the Clausius-Clapeyron equation relates the temperature and vapor pressure of a substance during a phase change. It is represented as \[\frac{dP}{dT} = \frac{\Delta H_{vap}}{T \cdot \Delta V}\] However, assuming ideal behavior where the volume of liquid is negligible compared to the volume of gas, the equation simplifies to\[\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] where \(\Delta H_{vap}\) is the molar enthalpy of vaporisation, \(R\) is the ideal gas constant, and \(T\) is temperature and \(P\) is the vapor pressure at points 1 and 2 respectively.

Convert Enthalpy of Vaporization

The enthalpy of vaporization given is in kJ/mol, but the gas constant is in J/K\cdot mol. Convert \(\Delta H_{vap}\) to J/mol to use with the gas constant \(R\).\[\Delta H_{vap} = 40.67 \text{ kJ/mol} = 40670 \text{ J/mol}\]

Apply the Clausius-Clapeyron Equation

Use the boiling point of water at standard pressure as \(T_1\) and the given temperature as \(T_2\). Note that the boiling point needs to be in Kelvin. Set up the equation with the known values and solve for \(T_2\).\[\ln\left(\frac{19}{760}\right) = \frac{40670}{8.3}\left(\frac{1}{373} - \frac{1}{T_2}\right)\] Our goal now is to solve for \(T_2\) (the unknown temperature).

Simplify the Equation

Divide both sides of the equation by the enthalpy over the gas constant to isolate the temperature term. Then, solve for \(1/T_2\), and finally, \(T_2\).\[\ln\left(\frac{19}{760}\right) \frac{8.3}{40670} = \frac{1}{373} - \frac{1}{T_2}\]

Calculate the Temperature

Convert the natural logarithm into a decimal using provided log values, perform the arithmetic, and finally calculate \(T_2\).\[\ln\left(\frac{19}{760}\right) = \ln\left(\frac{1}{40}\right) \approx \ln(2^{-5.32}) = -5.32 \cdot \ln(2) = -5.32 \cdot 0.3\] Next, plug this value back into the equation and solve for \(T_2\).

Give the Answer

Using the values from the previous steps, do the necessary calculations to find the numerical value of \(T_2\). Make sure your final answer corresponds to one of the options given in the exercise.

Key concepts

Vapor Pressure

Vapor pressure is an essential concept in physical chemistry often encountered when studying the behavior of liquids and gases. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid form. In simple terms, it is the pressure created by the molecules that escape from a liquid or solid and enter the gaseous phase, at a given temperature.

Equilibrium is key here; it means the rate at which the molecules evaporate from the liquid (or sublimate from the solid) is equal to the rate at which they condense back into the liquid (or solid). The vapor pressure of a substance is dependent on the temperature—the higher the temperature, the more molecules have enough kinetic energy to escape into the gas phase, leading to a higher vapor pressure.

When the vapor pressure equals the external pressure, the liquid will boil. This is why water boils at lower temperatures at high altitudes, where atmospheric pressure is lower, and why pressure cookers can cook food faster - by increasing the pressure, the boiling point of water is raised.

Enthalpy of Vaporization

Moving on to the enthalpy of vaporization, this term describes the amount of energy required to convert one mole of a substance from a liquid to a gas at constant pressure. It is a form of latent heat because it is energy used in a phase change process without changing the temperature of the substance. The energy essentially breaks the intermolecular forces that hold the liquid molecules together.

The enthalpy of vaporization ( ) is significant in calculations involving heat transfer during boiling or condensing processes. For instance, when water boils, it absorbs heat from its surroundings, which is then used to break the hydrogen bonds between the water molecules so they can escape into the air as gas. This heat gets 'hidden' in the vapor as potential energy. When the vapor condenses back into a liquid, the stored energy is released as sensible heat.

In the context of the exercise, the enthalpy of vaporization of water is given, and it plays a crucial role in applying the Clausius-Clapeyron equation to determine the temperature at which water has a specific vapor pressure.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationship between chemical reactions and energy changes involving heat. It provides a framework for understanding how energy is absorbed or released during chemical processes and how it affects matter's physical state. The laws of thermodynamics govern these energy transformations and stipulate, for instance, that energy cannot be created or destroyed (first law) and that the total entropy of an isolated system can never decrease over time (second law).

One key aspect of this field is studying the behavior of substances as they change phases, like from liquid to gas. This is highly relevant to the Clausius-Clapeyron equation, which describes how the temperature and pressure of a substance are related during a phase transition. Understanding the principles of thermodynamics is also crucial for solving problems related to heat, work, temperature, and thermodynamic functions such as entropy and free energy—concepts pivotal in determining the conditions for spontaneous chemical reactions and phase transitions.

Physical Chemistry for JEE

The Joint Entrance Examination (JEE) is a highly competitive exam for students aspiring to pursue engineering programs at premier institutes in India. Physical chemistry, one of the three pillars along with organic and inorganic chemistry, is an important subject covered in the JEE syllabus. Concepts like vapor pressure, enthalpy of vaporization, and thermodynamics are crucial topics within physical chemistry for JEE.

Mastery of these concepts is not just about memorization but also about understanding their applications, such as in the Clausius-Clapeyron equation that connects vapor pressure and temperature. The ability to apply these principles to solve complex problems is a critical skill assessed in the JEE. This includes understanding phase diagrams, interpreting the behavior of real gases as opposed to ideal gases, and solving quantitative problems using equations that describe physical phenomena. For anyone preparing for the JEE, it is important to have a strong grasp of these fundamental principles and to practice solving various types of problems.