Question

If \(10^{-4} \mathrm{dm}^{3}\) of water is introduced into a \(1.0 \mathrm{dm}^{3}\) flask at \(300 \mathrm{~K}\), how many moles of water are in the vapour phase when equilibrium is established? (Given : Vapour pressure of \(\mathrm{H}_{2} \mathrm{O}\) at \(300 \mathrm{~K}\) is \(3170 \mathrm{~Pa} ; R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(1.27 \times 10^{-3}\) mole (b) \(5.56 \times 10^{-3}\) mole (c) \(1.53 \times 10^{-2}\) mole (d) \(4.46 \times 10^{-2}\) mole

Short answer

1.27 * 10⁻³ mole

Step-by-step solution

Understand the given data

Water is introduced into a flask with a volume of 1.0 dm³ at a temperature of 300 K. The vapour pressure of H₂O at this temperature is 3170 Pa. The ideal gas constant R is 8.314 J K⁻¹ mol⁻¹.

State the ideal gas law

The number of moles of gas can be found using the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Calculate the number of moles

Rearrange the ideal gas law to solve for n: n = PV / (RT). Substitute the known values into the equation: n = (3170 Pa * 0.001 dm³) / (8.314 J K⁻¹ mol⁻¹ * 300 K).

Convert units

Convert the volume from dm³ to m³ because the pressure is given in Pa (1 dm³ = 0.001 m³). This step has already been performed in the previous step, where we converted 0.001 dm³ to m³.

Perform the calculation

n = (3170 Pa * 0.001 m³) / (8.314 J K⁻¹ mol⁻¹ * 300 K) = (3.170 J) / (2494.2 J mol⁻¹) = 1.27 * 10⁻³ mol.

Choose the correct answer

The calculated number of moles is 1.27 * 10⁻³. Thus, the answer is (a) 1.27 * 10⁻³ mole.

Key concepts

Vapour Pressure

Vapour pressure is a fundamental property of a liquid, referring to the pressure exerted by a vapor that is in thermodynamic equilibrium with its liquid phase at a given temperature. In simpler terms, when a liquid is placed in a closed container, some of the liquid will evaporate forming a vapor above it. As time passes, the rate of evaporation and the rate of condensation (vapor turning back to liquid) will become equal, reaching a state of equilibrium. At this point, the pressure exerted by the vapor is known as the vapour pressure.

Vapour pressure is critical because it is indicative of a liquid's evaporation rate and is also a function of temperature: as the temperature increases, vapour pressure increases too. Understanding vapour pressure is vital for solving a variety of problems in physical chemistry and for those preparing for competitive exams like JEE.

Ideal Gas Equation

The ideal gas equation is a cornerstone of gas laws, used to relate the pressure, volume, temperature, and number of moles of a gas. It is represented by the formula: \[ PV = nRT \]where P stands for pressure, V stands for volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in kelvin (K). This equation assumes that the gas behaves ideally, meaning the particles do not attract or repel each other and occupy no volume.

Despite its simplicity, the ideal gas law is incredibly powerful, allowing us to make predictions about the behavior of gases under different conditions. It's particularly useful in laboratory settings and is an essential concept in physical chemistry, forming the basis for more complex gas laws.

Molar Calculations

Molar calculations are a central aspect of chemistry, providing a bridge between the macroscopic world we observe and the microscopic world of atoms and molecules. The mole is the unit that allows chemists to count particles using a scale that's relevant to laboratory quantities of substances.

To perform molar calculations, one has to understand Avogadro's number (approximately \(6.022 \times 10^{23}\)), which defines the number of units in one mole of a substance. These calculations involve converting between mass, moles, and the number of particles. A foundational equation used in molar calculations is derived from the ideal gas law, which allows you to calculate the number of moles of gas from its pressure, volume, and temperature.

Physical Chemistry for JEE

As a competitive entry exam for engineering colleges in India, the Joint Entrance Examination (JEE) places a significant emphasis on physical chemistry. This branch of chemistry deals with understanding the physical properties of molecules, the forces that act upon them, and the characteristics of their chemical reactions.

The topics covered in physical chemistry for JEE include thermodynamics, chemical kinetics, chemical equilibrium, and solutions, among others. Grasping these concepts requires not only memorization but also a deep understanding of the underlying principles. They often involve mathematical problems, like the vapour pressure calculations or ideal gas law problems seen in our textbook exercise, which aim to strengthen a student's problem-solving skills and conceptual knowledge required for success in the JEE and future studies in engineering or physical sciences.