INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN
INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN
8th Edition
ISBN: 9780134580289
Author: CORWIN
Publisher: PEARSON
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Chapter 10, Problem 10E
Interpretation Introduction

(a)

Interpretation:

Change in the pressure (P) when volume (V) of a gas is decreased is to be stated.

Concept introduction:

To study the molecules in gaseous state different laws have been given by different scientists which relates different variables to each other. Boyle’s law which relates pressure and temperature states that when temperature is kept constant for a gas then the pressure and temperature follows the inverse relation. According to Boyle’s law, relation between pressure and volume is written below.

P1V, T is constantP=kVPV=k

Expert Solution
Check Mark

Answer to Problem 10E

Pressure increases with decrease in volume.

Explanation of Solution

According to the equation of Boyle’s law if volume is decreased for a particular or constant temperature then the pressure of the gas will be increased to keep the PV constant Boyle’s law for the gaseous molecule is stated as below.

P1V, T is constantP=kVPV=kP1V1=P2V2

k is a proportionality constant

P1 is the initial pressure of the gas.

P2 is the final pressure of the gas.

V1 is the initial volume of the container or gas

V2 is the final volume of the container or gas

Therefore, net value of pressure increases with decrease in volume.

Conclusion

Net value of pressure increases with decrease in volume.

Interpretation Introduction

(b)

Interpretation:

Change in the pressure (P) when temperature (T) of a gas is decreased is to be stated.

Concept introduction:

To study the molecules in gaseous state different laws have been given by different scientists which relates different variables to each other. Gay Lussac’s law one of the gas laws relate the pressure and temperature for a particular gas when rest of the variables is constant. Gay Lussac’s law or relation between pressure and temperature is written below.

PT, V is constantP=kT,V is constantPT=k

Expert Solution
Check Mark

Answer to Problem 10E

The pressure values decreases with decrease in the temperature of the gas.

Explanation of Solution

Relation given by Gay Lussac’s for the gaseous molecule is stated as below.

PT, V is constantP=kT,V is constantPT=kP1T1=P2T2

k is a proportionality constant

P1 is the initial pressure of the gas.

P2 is the final pressure of the gas.

T1 is the initial temperature of the gas

T2 is the final temperature of the gas

Ratio of pressure and temperature is constant. Pressure will be decreased with the decrease in value of temperature to keep the ratio PT constant.

Therefore, net value of pressure decreases with decrease in volume.

Conclusion

Pressure of a gas decreases with decrease in temperature.

Interpretation Introduction

(c)

Interpretation:

Change in the pressure when moles of a gas are decreased is to be stated.

Concept introduction:

To study the molecules in gaseous state different laws have been given by different scientists which relates different variables to each other. Avogadro’s law relates volume of a gas with the number of moles or molecules of the same gas. According to this law when temperature and pressure of a gas is kept constant then the equal volumes of gas contains equal number of molecules of that gas. Combined gas law which is the result of Boyle’s law, Charles’s law is, Gay Lussac’s law and Avogadro’s law is written below.

PT, V is constantP1V,T is constantVn, (P,T are constant)Pn×TV.

Expert Solution
Check Mark

Answer to Problem 10E

Net value of pressure decreases with decrease in number of moles.

Explanation of Solution

Relation given by all the variables of a gas is stated as below.

Pn×TVP=n×T×RV

Gas law is written below.

PV=nRT …(1)

Gas law for two different gases is shown below.

P1×V1n1×T1×R=P2×V2n2×T2×R …(2)

At constant volume and temperature, equation (2) is shown below.

P1n1=P2n2

Where,

P1 is the initial pressure exerted by gas molecules.

P2.is the final pressure exerted by gas molecules.

n1 is the initial number of moles of gas.

n2 is final number of moles of a gas.

According to the relation given in equation (1), pressure is directly proportional to the number of moles. Pressure will be decreased with the decrease in number of moles to maintain the gas law when all other variables are kept constant.

Therefore, pressure of a gas decreases with decrease in number of moles.

Conclusion

Net value of pressure decreases with decrease in number of moles (n) of a gas.

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Chapter 10 Solutions

INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN

Ch. 10 - Prob. 1KTCh. 10 - Prob. 2KTCh. 10 - Prob. 3KTCh. 10 - Prob. 4KTCh. 10 - Prob. 5KTCh. 10 - Prob. 6KTCh. 10 - Prob. 7KTCh. 10 - Prob. 8KTCh. 10 - Prob. 9KTCh. 10 - Prob. 10KTCh. 10 - Prob. 11KTCh. 10 - Prob. 12KTCh. 10 - Prob. 13KTCh. 10 - Prob. 14KTCh. 10 - Prob. 15KTCh. 10 - Prob. 16KTCh. 10 - Prob. 17KTCh. 10 - Prob. 18KTCh. 10 - Prob. 19KTCh. 10 - Prob. 20KTCh. 10 - Prob. 21KTCh. 10 - Prob. 22KTCh. 10 - Prob. 23KTCh. 10 - Prob. 24KTCh. 10 - Prob. 1ECh. 10 - Prob. 2ECh. 10 - Prob. 3ECh. 10 - Prob. 4ECh. 10 - Prob. 5ECh. 10 - Prob. 6ECh. 10 - Prob. 7ECh. 10 - Prob. 8ECh. 10 - Prob. 9ECh. 10 - Prob. 10ECh. 10 - Prob. 11ECh. 10 - Prob. 12ECh. 10 - Prob. 13ECh. 10 - Prob. 14ECh. 10 - Prob. 15ECh. 10 - Prob. 16ECh. 10 - Prob. 17ECh. 10 - Prob. 18ECh. 10 - Prob. 19ECh. 10 - Prob. 20ECh. 10 - Prob. 21ECh. 10 - Prob. 22ECh. 10 - Prob. 23ECh. 10 - Prob. 24ECh. 10 - Prob. 25ECh. 10 - Prob. 26ECh. 10 - Prob. 27ECh. 10 - Prob. 28ECh. 10 - Prob. 29ECh. 10 - Prob. 30ECh. 10 - Prob. 31ECh. 10 - Prob. 32ECh. 10 - Prob. 33ECh. 10 - Prob. 34ECh. 10 - Prob. 35ECh. 10 - Prob. 36ECh. 10 - Prob. 37ECh. 10 - Prob. 38ECh. 10 - Prob. 39ECh. 10 - Prob. 40ECh. 10 - Prob. 41ECh. 10 - Prob. 42ECh. 10 - Prob. 43ECh. 10 - Prob. 44ECh. 10 - Prob. 45ECh. 10 - Prob. 46ECh. 10 - Prob. 47ECh. 10 - Prob. 48ECh. 10 - Prob. 49ECh. 10 - Prob. 50ECh. 10 - Prob. 51ECh. 10 - Prob. 52ECh. 10 - Prob. 53ECh. 10 - Prob. 54ECh. 10 - Prob. 55ECh. 10 - Prob. 56ECh. 10 - Prob. 57ECh. 10 - Prob. 58ECh. 10 - Prob. 59ECh. 10 - Prob. 60ECh. 10 - Prob. 61ECh. 10 - Prob. 62ECh. 10 - Prob. 63ECh. 10 - Prob. 64ECh. 10 - Prob. 65ECh. 10 - Prob. 66ECh. 10 - Prob. 67ECh. 10 - Prob. 68ECh. 10 - Prob. 69ECh. 10 - Prob. 70ECh. 10 - Prob. 71ECh. 10 - Prob. 72ECh. 10 - Prob. 73ECh. 10 - Prob. 74ECh. 10 - Prob. 75ECh. 10 - Prob. 76ECh. 10 - Prob. 77ECh. 10 - Prob. 78ECh. 10 - Prob. 79ECh. 10 - Prob. 80ECh. 10 - Prob. 81ECh. 10 - Prob. 82ECh. 10 - Prob. 83ECh. 10 - Prob. 84ECh. 10 - Prob. 1STCh. 10 - Prob. 2STCh. 10 - Prob. 3STCh. 10 - Prob. 4STCh. 10 - Prob. 5STCh. 10 - Prob. 6STCh. 10 - Prob. 7STCh. 10 - Prob. 8STCh. 10 - Prob. 9STCh. 10 - Prob. 10STCh. 10 - Prob. 11STCh. 10 - Prob. 12STCh. 10 - Prob. 13STCh. 10 - Prob. 14STCh. 10 - Prob. 15STCh. 10 - Prob. 16STCh. 10 - Prob. 17STCh. 10 - Prob. 18STCh. 10 - Prob. 19ST
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