Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Question: The vapor pressure of water at 40.0ºC is 7.34×10³ N/m². Using the ideal gas law, calculate the density of water vapor in g/m³ that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature (51.1 g/m³).

Answer

The density of water vapor is 50.746 g/m³.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Defining the ideal gas law

Ideal gas law states that the product of pressure and volume of a gas is equal to the product of temperature in Kelvin and the universal gas constant. It is expressed as;

$P V = n R T$ ……………….(1)

Where is the pressure, is the volume, is the number of moles,is the universal gas constant, and is the temperature in Kelvin.

Step 3: Evaluate the given factors and convert them into suitable units

The value of pressure and temperature are given.

$P = 7.34 \times 10^{3} N / m^{3} T = 40.0 C = (40 + 273.15) K = 313.15 K$

The value of is 8.314 Jmol^-1K^-1. No of moles is the ratio of a given mass of a substance to the molar mass. It is expressed as;

$n = \frac{M a s s}{M o l a r M a s s}$ ……..…(2)

The molar mass of water is 18g. Substitute this in equation (2).

$n = \frac{M a s s}{18 g}$ …………….(3)

Step 3: Find density from the ideal gas formula

The density of water vapor can be derived from the ideal gas formula.

Substitute equation (3) in equation (1)

$P V = \frac{M a s s}{18 g} R T \frac{M a s s}{V} = \frac{P \times 18 g}{R T}$

The ratio of mass to volume is the density. Therefore,

$D e n s i t y = \frac{P \times 18 g}{R T} = \frac{7.34 \times 10^{3} N / m^{2} \times 18 g}{8.314 J m o l^{- 1} K^{- 1} \times 313.15 K} = 50.746 g / m^{3}$

This value is approximately equal to 51.1 g/m³, which is the saturation vapor density at the given temperature. The small variation is due to the error caused by rounding the values.

So, the calculated vapor density is equal to 50.746 g/m³.

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College Physics (Urone)

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Short Answer

Question: The vapor pressure of water at 40.0ºC is 7.34×10³ N/m². Using the ideal gas law, calculate the density of water vapor in g/m³ that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature (51.1 g/m³).

Step by Step Solution

Step 1: Defining the ideal gas law

Step 3: Evaluate the given factors and convert them into suitable units

Step 3: Find density from the ideal gas formula

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College Physics (Urone)

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Short Answer

Question: The vapor pressure of water at 40.0ºC is 7.34×103 N/m2. Using the ideal gas law, calculate the density of water vapor in g/m3 that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature (51.1 g/m3).

Step by Step Solution

Step 1: Defining the ideal gas law

Step 3: Evaluate the given factors and convert them into suitable units

Step 3: Find density from the ideal gas formula

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Question: The vapor pressure of water at 40.0ºC is 7.34×10³ N/m². Using the ideal gas law, calculate the density of water vapor in g/m³ that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature (51.1 g/m³).