/*! 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 14 Calculate the volume of \(1.00 \... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the volume of \(1.00 \mathrm{~mol}\) of liquid water at a temperature of \(20^{\circ} \mathrm{C}\) (at which its density is \(998 \mathrm{~kg} / \mathrm{m}^{3}\) ), and compare this volume with the volume occupied by \(1.00 \mathrm{~mol}\) of steam at \(200^{\circ} \mathrm{C}\). Assume the steam is at atmospheric pressure and can be treated as an ideal gas.

Short Answer

Expert verified
The volume of 1 mol of steam is about 2160 times greater than that of liquid water at 20°C.

Step by step solution

01

Find the Volume of Liquid Water

To find the volume of liquid water, we use the density-volume-mass relationship: \( \text{density} = \frac{\text{mass}}{\text{volume}} \). The density of water is given as \( 998 \, \text{kg/m}^3 \). We need to find the volume corresponding to \(1.00 \, \text{mol}\) of water.First, we convert the moles to mass using the molar mass of water \((18.02 \, \text{g/mol})\):\[ \text{mass} = 1.00 \, \text{mol} \times 18.02 \, \text{g/mol} = 18.02 \, \text{g} = 0.01802 \, \text{kg} \]Then, we calculate the volume:\[ \text{volume} = \frac{\text{mass}}{\text{density}} = \frac{0.01802 \, \text{kg}}{998 \, \text{kg/m}^3} = 1.8056 \times 10^{-5} \, \text{m}^3 \]
02

Calculate the Volume of Steam using the Ideal Gas Law

For the steam, we assume it behaves as an ideal gas. We use the ideal gas law: \( PV = nRT \).We need to calculate the volume \(V\) for \( n = 1.00 \, \text{mol} \). The ideal gas constant \(R\) is \(8.314 \, \text{J/mol·K}\), and the atmospheric pressure \(P\) is \(101325 \, \text{Pa}\). The temperature \(T\) should be in Kelvin:\[ T = 200^{\circ} \text{C} + 273.15 = 473.15 \, \text{K} \]Using these values, we solve for volume \(V\):\[ V = \frac{nRT}{P} = \frac{(1.00 \, \text{mol})(8.314 \, \text{J/mol·K})(473.15 \, \text{K})}{101325 \, \text{Pa}} = 0.03898 \, \text{m}^3 \]
03

Compare the Volumes of Liquid Water and Steam

We have calculated the volume of \(1.00 \, \text{mol}\) of liquid water to be \(1.8056 \times 10^{-5} \, \text{m}^3\) and the volume of the steam to be \(0.03898 \, \text{m}^3\).To compare, note that the volume of steam is significantly larger than the volume of liquid water. Specifically, \(0.03898 \, \text{m}^3\) is approximately \(2160\) times bigger than \(1.8056 \times 10^{-5} \, \text{m}^3\).

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Key Concepts

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

Volume Calculation
In the given exercise, our goal is to compute the volume of both liquid water and steam. When determining volume, it's crucial to understand the relationship between density, mass, and volume. This is expressed in the formula:
  • \( \text{density} = \frac{\text{mass}}{\text{volume}} \)
  • Therefore, \( \text{volume} = \frac{\text{mass}}{\text{density}} \)
For the liquid water, knowing the mass in kilograms and the given density, you can plug these values into the formula to find the volume. Moreover, when dealing with gases (like steam), the Ideal Gas Law helps us calculate volume based on pressure, temperature, and amount of substance in moles. This law is given by:
  • \( PV = nRT \)
  • Solving for \( V \) gives \( V = \frac{nRT}{P} \)
When you apply these formulas correctly, you can compare how vastly different the volumes of liquids and gases are under similar conditions.
Density of Water
Density is a measure of how much mass fits into a given volume. For water at \(20^{\circ} \mathrm{C}\), this is \(998 \mathrm{~kg}/ \mathrm{m}^{3} \). Density plays a critical role in scuba diving, cooking, and various scientific calculations, like our exercise here.
To understand density, think about how tightly packed the molecules are within the substance. When calculating volume, you need the mass, which you often get from the molecular makeup of the compound. For water, knowing that the density is \(998 \mathrm{~kg}/ \mathrm{m}^{3} \) allows you to calculate the volume for any given mass by rearranging the formula, \( \text{density} = \frac{\text{mass}}{\text{volume}} \), to find volume.
This basic understanding of density is also applicable to other substances, each having its own unique density value which can change with temperature or pressure. This concept is fundamental for fields ranging from hydrology to material science.
Molar Mass of Water
The molar mass of a compound is the mass of one mole of its molecules. For water \( \text{H}_2\text{O} \), the molar mass is \(18.02 \text{g/mol}\). This value comes from the sum of the atomic masses of hydrogen and oxygen:
  • Each hydrogen atom has a mass of about \(1.01 \text{g/mol}\), and there are two hydrogen atoms: \(2 \times 1.01 = 2.02 \text{g/mol} \).
  • One oxygen atom has a mass of about \(16 \text{g/mol}\).
  • Adding them together gives \(18.02 \text{g/mol}\).
In calculations, this molar mass helps convert between the number of moles and the mass in grams. For example, to find the mass of one mole of water, you multiply the molar mass \(18.02 \text{g/mol}\) by the number of moles. This is fundamental for stoichiometry in chemistry, helping you predict how substances will react and in what amounts.
Temperature Conversion
Temperature conversion is essential in chemistry, particularly when applying the Ideal Gas Law, where temperatures must be in Kelvin. Kelvin is the standard unit of temperature in scientific calculations.
To convert from Celsius to Kelvin, you simply add \(273.15\). For instance, \(20^{\circ} \mathrm{C}\) becomes:
  • \(20^{\circ} \text{C} + 273.15 = 293.15 \text{K} \)
And \(200^{\circ} \text{C}\) becomes:
  • \(200^{\circ} \text{C} + 273.15 = 473.15 \text{K} \)
This conversion is necessary because the Ideal Gas Law, \( PV = nRT \), uses Kelvin to ensure that the temperature scale starts at absolute zero, making the mathematical relationships valid. Understanding this concept is important for consistency and accuracy in scientific experiments and calculations.

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Most popular questions from this chapter

A cylinder with a piston contains \(0.250 \mathrm{~mol}\) of ideal oxygen at a pressure of \(2.40 \times 10^{5} \mathrm{~Pa}\) and a temperature of \(355 \mathrm{~K}\). The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a \(p V\) diagram. (b) Compute the temperature during the isothermal compression. (c) Compute the maximum pressure. (d) Compute the total work done by the piston on the gas during the series of processes.

An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

A \(20.0 \mathrm{~L}\) tank contains \(0.225 \mathrm{~kg}\) of helium at \(18.0^{\circ} \mathrm{C}\). The molar mass of helium is \(4.00 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

(a) How much heat does it take to increase the temperature of 2.50 moles of an ideal monatomic gas from \(25.0^{\circ} \mathrm{C}\) to \(55.0^{\circ} \mathrm{C}\) if the gas is held at constant volume? (b) How much heat is needed if the gas is diatomic rather than monatomic? (c) Sketch a \(p V\) diagram for these processes.

Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of \(32.0 \mathrm{~g} / \mathrm{mol}\). (a) What is the root-mean-square speed of an oxygen molecule at a temperature of \(300 \mathrm{~K} ?\) (b) What is its average translational kinetic energy at that speed?

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