/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Pure water vapour is trapped in ... [FREE SOLUTION] | 91Ó°ÊÓ

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Pure water vapour is trapped in a vessel of volume \(10 \mathrm{~cm}^{3}\). The relative humidity is \(40 \%\). The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

Short Answer

Expert verified
The vapor starts condensing at a volume of 4 cm³.

Step by step solution

01

Understand the Concept of Relative Humidity

Relative humidity is the ratio of the partial pressure of water vapor to the equilibrium vapor pressure of water at a given temperature. When it is 100%, the vapor is saturated and can start to condense. In this problem, the relative humidity is 40%.
02

Calculate the Partial Pressure of Water Vapor

To begin, let's denote the saturation vapor pressure at the given temperature as \( P_{sat} \). Since the relative humidity is 40%, the partial pressure of the vapor \( P_{vapor} \) will be \( P_{vapor} = 0.4 \times P_{sat} \).
03

Determine the Volume When Vapor Starts to Condense

Since vapor will start condensing when the partial pressure reaches its saturation value (100% relative humidity), we need to determine the volume when this happens. Using Boyle’s Law, \( P_1V_1 = P_2V_2 \), where \( P_1 = 0.4P_{sat}, V_1 = 10 \: cm^3 \) and \( P_2 = P_{sat} \), we find \( V_2 \).
04

Apply Boyle’s Law to Find New Volume

Using \( P_1 = 0.4P_{sat} \) and \( P_2 = P_{sat} \), substitute into Boyle's Law: \( 0.4P_{sat} \times 10 = P_{sat} \times V_2 \). Simplifying gives \( V_2 = \frac{0.4 \times 10}{1} = 4 \: cm^3 \).
05

Conclusion

The vapor in the vessel will start to condense when compressed isothermally to a volume of \(4 \: cm^3\), as the pressure will reach the saturation vapour pressure at this volume.

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Key Concepts

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

Partial Pressure
Partial pressure is a key concept in understanding how gases behave in a mixture. Particularly, it tells us how much pressure is exerted by one type of gas within a mixture of gases. The partial pressure of water vapor in our scenario relates directly to relative humidity, which measures how close the air is to being saturated with water vapor. If 100% is reached, the air is fully saturated, and condensation can occur.

To find the partial pressure, we use the formula:
  • Partial pressure of water vapor, denoted as \( P_{vapor} \), is given by \( P_{vapor} = \text{relative humidity} \times P_{sat} \).
This equation tells us that the partial pressure is a fraction of the saturation vapor pressure, \( P_{sat} \), which is the pressure exerted by a vapor in thermodynamic equilibrium with its liquid phase at a given temperature. In our problem, with 40% relative humidity, the calculation becomes \( P_{vapor} = 0.4 \times P_{sat} \). Understanding this relationship helps in predicting at what point a vapor will begin to condense.
Vapor Condensation
Vapor condensation is the process where water vapor (or any gas) changes to its liquid form. This process begins when the vapor becomes saturated, meaning it reaches a relative humidity of 100%. Once the saturation point is achieved, any further compression of the gas or cooling will lead to condensation.

In the example of an enclosed vapor volume, when the vapor is compressed, its partial pressure increases. The moment this pressure hits the saturation vapor pressure, the vapor starts to condense.

To predict when condensation will occur, we adjust the volume using Boyle's Law, checking when the partial pressure of the vapor equals the saturation pressure. Condensation disrupts the vapor-liquid equilibrium and is crucial for understanding natural processes like weather patterns and industrial applications such as distillation.
Boyle's Law
Boyle's Law is a fundamental principle in chemistry describing how a constant temperature affects the behavior of gases. The law is mathematically expressed as \( P_1V_1 = P_2V_2 \), where \( P \) is the pressure, \( V \) is the volume, and the subscripts 1 and 2 represent the initial and final states of the gas, respectively.

According to Boyle's Law, for a given mass of confined gas at constant temperature, the product of its pressure and volume is always constant. In other words, if the volume decreases, the pressure increases, and vice versa, assuming temperature remains stable.

In our particular case, starting with a 40% saturated vapor at a volume of 10 cm³, compressing the vapor to a smaller volume increases its pressure until it equals the saturation pressure, causing condensation. This application of Boyle's Law provides valuable insights into the behavior of real gases during physical processes such as thermal dynamics, engineering, and weather prediction.

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Most popular questions from this chapter

A container of volume 50 cc contains air (mean molecular weight \(=28 \cdot 8 \mathrm{~g}\) ) and is open to atmosphere where the pressure is \(100 \mathrm{kPa}\). The container is kept in a bath containing melting ice \(\left(0^{\circ} \mathrm{C}\right)\). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water \(\left(100^{\circ} \mathrm{C}\right)\). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.

A faulty barometer contains certain amount of air and saturated water vapour. It reads \(74 \cdot 0 \mathrm{~cm}\) when the atmospheric pressure is \(76.0 \mathrm{~cm}\) of mercury and reads \(72 \cdot 10 \mathrm{~cm}\) when the atmospheric pressure is \(74 \cdot 0 \mathrm{~cm}\) of mercury. Saturation vapour pressure at the air temperature \(=1 \cdot 0 \mathrm{~cm}\) of mercury. Find the length of the barometer tube above the mercury level in the reservoir.

A bucket full of water is placed in a room at \(15^{\circ} \mathrm{C}\) with initial relative humidity \(40 \%\). The volume of the room is \(50 \mathrm{~m}^{3}\). (a) How much water will evaporate ? (b) If the room temperature is increased by \(5^{\circ} \mathrm{C}\), how much more water will evaporate ? The saturation vapour pressure of water at \(15^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\) are \(1.6 \mathrm{kPa}\) and \(2 \cdot 4 \mathrm{kPa}\) respectively.

The average translational kinetic energy of air molecules is \(0.040 \mathrm{eV} \quad\left(1 \mathrm{eV}=1 \cdot 6 \times 10^{-19} \mathrm{~J}\right) . \quad\) Calculate the temperature of the air. Boltzmann constant \(k=1 \cdot 38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\)

A vessel of volume \(V_{0}\) contains an ideal gas at pressure \(p_{0}\) and temperature \(T\). Gas is continuously pumped out of this vessel at a constant volume-rate \(d V / d t=r\) keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.

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