/*! 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 88 The vapor pressure is the pressu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The vapor pressure is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The relative humidity is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same terperature, expressed as a percentage. The air is saturated when the humidity is 100\(\%\) . (a) The vapor pressure of water at \(20.0^{\circ} \mathrm{C}\) is \(2.34 \times 10^{3} \mathrm{Pa}\) . If the air temperature is \(20.0^{\circ} \mathrm{C}\) and the relative humidity is 60\(\%\) what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 \(\mathrm{m}^{3}\) of air? (The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) . Assume that water vapor can be treated as an ideal gas.)

Short Answer

Expert verified
(a) 1404 Pa; (b) 10.38 g/m³.

Step by step solution

01

Calculate the Partial Pressure of Water Vapor

Since the relative humidity (RH) is given as 60% when the vapor pressure at 20.0°C is 2340 Pa, the partial pressure of water vapor can be found using the formula: \( P_{w} = \text{RH} \times P_0 \), where:- \( P_{w} \) is the partial pressure of water vapor,- \( \text{RH} = 0.60 \), - \( P_0 = 2340 \text{ Pa} \).Therefore, \( P_{w} = 0.60 \times 2340 = 1404 \text{ Pa} \).
02

Use Ideal Gas Law to Find Density of Water Vapor

To determine the mass of water in 1.00 m³ of air, use the ideal gas law equation: \( PV = nRT \). Rearrange to find \( n \), the amount of substance (in moles): \( n = \frac{PV}{RT} \).Given:- Pressure \( P_{w} = 1404 \text{ Pa} \),- Volume \( V = 1.00 \text{ m}^3 \),- Ideal gas constant \( R = 8.314 \text{ J/mol·K} \),- Temperature \( T = 293.15 \text{ K} \) (which is 20°C in absolute temperature).So, \( n = \frac{1404 \times 1}{8.314 \times 293.15} = 0.5767 \text{ moles} \).
03

Calculate the Mass of Water in the Air

To convert the amount of substance in moles to mass, use the formula: \( m = n \times M \), where:- \( m \) is the mass,- \( n = 0.5767 \text{ moles} \),- Molar Mass \( M = 18.0 \text{ g/mol} \).Thus, \( m = 0.5767 \times 18.0 = 10.38 \text{ g} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Relative Humidity
Relative humidity is an important measurement in meteorology and atmospheric science. It refers to the amount of moisture present in the air compared to the maximum amount of water vapor that air could hold at a given temperature. This value is expressed as a percentage. A relative humidity of 50% means the air holds half of the water vapor needed for it to be fully saturated at that temperature.

Understanding relative humidity helps us predict weather patterns and human comfort levels. High humidity can make temperatures feel hotter, while low humidity can make the air feel cooler. Relative humidity is calculated using the formula:
\( \text{Relative Humidity} = \frac{\text{Partial Pressure of Water Vapor}}{\text{Vapor Pressure of Water}} \times 100 \% \).

In the exercise, when the relative humidity is 60%, it means the air contains 60% of the water vapor it could hold at that temperature. Calculating relative humidity involves knowing both the partial pressure of water vapor in the air and the vapor pressure at that particular temperature.
Ideal Gas Law
The Ideal Gas Law is a useful equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas. It is usually expressed as:
\( PV = nRT \), where:
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume the gas occupies,
  • \(n\) is the number of moles of the gas,
  • \(R\) is the ideal gas constant, and
  • \(T\) is the temperature in Kelvin.
This equation assumes gases behave ideally, which works well under a range of conditions but can deviate at high pressures or low temperatures. For water vapor, assuming ideal behavior simplifies calculations, like in the exercise.

By rearranging the Ideal Gas Law, you can solve for any variable, such as determining the number of moles. This makes it a versatile tool in many scientific calculations, such as finding the mass of water vapor in air.
Partial Pressure
Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It's essential in understanding how different gases behave independently within a mix, such as atmospheric air, which contains nitrogen, oxygen, water vapor, and other gases.

The concept of partial pressures is part of Dalton's Law of Partial Pressures, which states that the total pressure exerted by a mixture is the sum of the partial pressures of the individual gases present.
\( P_{\text{total}} = P_{1} + P_{2} + P_{3} + ... \)

In the exercise, we calculated the partial pressure of water vapor using the given relative humidity. This value represents the contribution of water vapor to the total atmospheric pressure. Knowing the partial pressure is crucial for determining humidity levels and for further calculations using the Ideal Gas Law.
Phase Equilibrium
Phase equilibrium occurs when a substance is in a state where its various phases—vapor, liquid, or solid—exist together without any net change over time. At equilibrium, the rate of evaporation equals the rate of condensation in a closed system, like the point where vapor pressure is measured.

This concept is key to understanding vapor pressure itself. Vapor pressure reflects the tendency of molecules to escape from a liquid or solid into the gaseous phase. When these phases are in dynamic balance, the vapor pressure of the system is constant if the temperature remains stable.

For instance, the vapor pressure of water at \(20.0^{\circ} \text{C}\) given in the problem is a characteristic value indicating phase equilibrium at this temperature. Understanding this equilibrium helps in predicting if condensation or evaporation will occur under changing environmental conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The speed of propagation of a sound wave in air at \(27^{\circ} \mathrm{C}\) is about 350 \(\mathrm{m} / \mathrm{s}\) . Calculate, for comparison, (a) \(v_{\mathrm{ms}}\) for nitrogen molecules and (b) the rms value of \(v_{x}\) at this temperature. The molar mass of nitrogen \(\left(\mathrm{N}_{2}\right)\) is \(28.0 \mathrm{g} / \mathrm{mol} .\)

A person at rest inhales 0.50 \(\mathrm{L}\) of air with each breath at a pressure of 1.00 atm and a temperature of \(20.0^{\circ} \mathrm{C}\) . The inhaled air is 21.0\(\%\) oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of \(2,000 \mathrm{m}\) but the temperature is still \(20.0^{\circ} \mathrm{C}\). Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report "shortness of breath" at high elevations.

In the ideal-gas equation, the number of moles per volume \(n / V\) is simply equal to \(p / R T\) . In the van der Waals equation, solving for \(n / V\) in terms of the pressure \(p\) and temperature \(T\) is somewhat more involved. (a) Show the van der Waals equation can be written as $$\frac{n}{V}=\left(\frac{p+a n^{2} / V^{2}}{R T}\right)\left(1-\frac{b n}{V}\right)$$ (b) The van der Waals parameters for hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{S}\right)\) are \(a=0.448 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.29 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol}\) . Determine the number of moles per volume of \(\mathrm{H}_{2} \mathrm{S}\) gas at \(127^{\circ} \mathrm{C}\) and an absolute pressure of \(9.80 \times 10^{5} \mathrm{Pa}\) as follows: (i) Calculate a first approximation using the ideal-gas equation, \(n / V=p / R T\) . (ii) Substitute this approximation for \(n / V\) into the right-hand side of the equation in part (a). The result is a new, improved approximation for \(n / V\) . (iii) Substitute the new approximation for \(n / V\) into the right-hand side of the equation in (a). The result is a further improved approximation for \(n / V\) . (iv) Repeat step (iii) until successive approximations agree to the desired level of accuracy (in this case, to three significant figures). (c) Compare your final result in part (b) to the result \(p / R T\) obtained using the ideal-gas equation. Which result gives a larger value of \(n / V ?\)

For carbon dioxide gas \(\left(\mathrm{CO}_{2}\right),\) the constants in the van der Waals equation are \(a=0.364 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.27 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (a) If 1.00 \(\mathrm{mol}\) of \(\mathrm{CO}_{2} \mathrm{gas}\) at 350 \(\mathrm{K}\) is confined to a volume of 400 \(\mathrm{cm}^{3}\) , find the pressure of the gas using the ideal-gas equation and the van der Waals equation. (b) Which equation gives a lower pressure? Why? What is the percentage difference of the van der Waals equation result from the ideat-gas equation result? (c) The gas is kept at the same temperature as it expands to a volume of 4000 \(\mathrm{cm}^{3} .\) Repeat the calculations of parts (a) and (b). (d) Explain how your calculations show that the van der Waals equation is equivalent to the ideat-gas equation if \(n / V\) is small.

Atmosphere of Titan. Titan, the largest satellite of Saturn, has a thick nitrogen atmosphere. At its surface, the pressure is 1.5 Earth-atmospheres and the temperature is 94 \(\mathrm{K}\) . (a) What is the surface temperature in \(^{\circ} \mathrm{C}\) ? (b) Calculate the surface density in Titar's atmosphere in molecules per cubic meter (c) Compare the density of Titan's surface atmosphere to the density of Earth's atmosphere at \(22^{\circ} \mathrm{C}\) . Which body has denser atmosphere?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.