EBK INTRODUCTORY CHEMISTRY
EBK INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 9780100480483
Author: DECOSTE
Publisher: YUZU
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Chapter 13, Problem 97QAP
Interpretation Introduction

Interpretation:

The volume of the mixture of gas at STP and partial pressure of each gas in the mixture at STP should be calculated.

Concept Introduction:

According to ideal gas equation:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The value of Universal gas constant can be taken as 0.082 L atm K1 mol1.

Mass of gas can be calculated from number of moles and molar mass as follows:

m=n×M

Here, n is number of moles, m is mass and M is molar mass.

Expert Solution & Answer
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Answer to Problem 97QAP

15.65 L, PO2=0.2237 atm, PN2=0.2556 atm, PCO2=0.1626 atm and PNe=0.3578 atm.

Explanation of Solution

Given Information:

The mass of oxygen gas, nitrogen gas, carbon dioxide gas and neon gas is 5 g.

First calculate the number of moles of each gas as follows:

n=mM

Molar mass of oxygen gas is 32 g/mol thus, number of moles will be:

nO2=5 g32 g/mol=0.1563 mol

Molar mass of nitrogen gas is 28 g/mol thus,

nN2=5 g28 g/mol=0.1786 mol

Molar mass of carbon dioxide is 44.01 g/mol thus,

nCO2=5 g44.01 g/mol=0.1136 mol

Molar mass of neon gas is 20.18 g/mol thus,

nNe=5 g20.18 g/mol=0.25 mol

Number of moles of mixture can be calculated as follows:

nmix=nO2+nN2+nCO2+nNe

Putting the values,

nmix=(0.1563+0.1786+0.1136+0.25) mol=0.6985 mol

At STP, the value of temperature is 273.15 K and pressure is 1 atm thus, volume can be calculated using the ideal gas equation as follows:

Vmix=nmixRTP

Putting the values,

V=(0.6985 mol)(0.082 L atm K1 mol1)(273.15 K)(1 atm)=15.65 L

Therefore, volume mixture of gases is 15.65 L.

Partial pressure of each gas at STP can be calculated considering the number of moles of individual gases using the ideal gas equation:

P=nRTV

Thus, partial pressure of oxygen gas can be calculated as follows:

PO2=(0.1563 mol)(0.082 L atm K1 mol1)(273.15 K)(15.65 L)=0.2237 atm

Partial pressure of nitrogen gas can be calculated as follows:

PN2=(0.1786 mol)(0.082 L atm K1 mol1)(273.15 K)(15.65 L)=0.2556 atm

Partial pressure of carbon dioxide gas can be calculated as follows:

PCO2=(0.1136 mol)(0.082 L atm K1 mol1)(273.15 K)(15.65 L)=0.1626 atm

Partial pressure of neon gas can be calculated as follows:

PNe=(0.25 mol)(0.082 L atm K1 mol1)(273.15 K)(15.65 L)=0.3578 atm.

Conclusion

Therefore, volume of mixture of gases is 15.65 L, partial pressure of oxygen, nitrogen, carbon dioxide and neon gas is 0.2237 atm, 0.2556 atm, 0.1626 atm and 0.3578 atm respectively.

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Chapter 13 Solutions

EBK INTRODUCTORY CHEMISTRY

Ch. 13.6 - Prob. 13.10SCCh. 13.8 - Prob. 1CTCh. 13.10 - trong>Exercise 13.11 Calculate the volume of...Ch. 13.10 - at if STP was defined as normal room temperature...Ch. 13.10 - Prob. 13.12SCCh. 13 - Prob. 1ALQCh. 13 - Prob. 2ALQCh. 13 - Prob. 3ALQCh. 13 - Prob. 4ALQCh. 13 - Prob. 5ALQCh. 13 - Prob. 6ALQCh. 13 - Prob. 7ALQCh. 13 - Prob. 8ALQCh. 13 - Prob. 9ALQCh. 13 - Prob. 10ALQCh. 13 - Prob. 11ALQCh. 13 - Prob. 12ALQCh. 13 - Prob. 13ALQCh. 13 - Draw molecular—level views than show the...Ch. 13 - Prob. 15ALQCh. 13 - Prob. 16ALQCh. 13 - Prob. 17ALQCh. 13 - Prob. 18ALQCh. 13 - Prob. 19ALQCh. 13 - Prob. 20ALQCh. 13 - You are holding two balloons of the same volume....Ch. 13 - Prob. 22ALQCh. 13 - Prob. 23ALQCh. 13 - The introduction to this chapter says that "we...Ch. 13 - Prob. 2QAPCh. 13 - Prob. 3QAPCh. 13 - Prob. 4QAPCh. 13 - Prob. 5QAPCh. 13 - Prob. 6QAPCh. 13 - Prob. 7QAPCh. 13 - Prob. 8QAPCh. 13 - Prob. 9QAPCh. 13 - Prob. 10QAPCh. 13 - Make the indicated pressure conversions....Ch. 13 - Prob. 12QAPCh. 13 - Prob. 13QAPCh. 13 - Prob. 14QAPCh. 13 - Prob. 15QAPCh. 13 - Prob. 16QAPCh. 13 - Prob. 17QAPCh. 13 - Prob. 18QAPCh. 13 - Prob. 19QAPCh. 13 - Prob. 20QAPCh. 13 - Prob. 21QAPCh. 13 - Prob. 22QAPCh. 13 - 3. 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