EBK CHEMISTRY:CENTRAL SCIENCE
EBK CHEMISTRY:CENTRAL SCIENCE
13th Edition
ISBN: 9780321924520
Author: Brown
Publisher: PEARSON CO
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Question
Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The pressure exerted by N2 gas by ideal gas equation.

(a)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The pressure exerted by N2 gas by ideal gas equation is 176.80atm .

Explanation of Solution

Given

The volume of the gas is 1100.0L .

The temperature of gas is 280°C .

The mass of N2 gas is 120.0kg .

The molar mass of nitrogen gas is 28.0g/mol .

The conversion kilogram (kg) into gram (g) is done as,

    1kg=103g

Hence, the conversion 120kg into gram is,

    120kg=120×103g

The conversion of temperature from °C into Kelvin is done as,

    T(K)=T(°C)+273

The conversion of temperature from 280°C into Kelvin is,

    T(K)=(280+273)K=553K

The pressure exerted by N2 gas is calculated by the formula,

    P=mRTMV

Where,

  • m is the mass of the gas.
  • M is the molar mass of the gas.
  • P is the absolute pressure of the gas.
  • V is the volume of the gas.
  • T is the absolute temperature of the gas.
  • R is the gas constant (0.08206Latm/molK) .
Substitute the given values of T,m,M,V and R in above equation.

    P=120× 103g×0.08206Latm/molK×553K28.0g/mol×1100.0L=176.80atm
Conclusion



The pressure exerted by N2 gas by ideal gas equation is 176.80atm .

(b)

Interpretation Introduction

To determine: The pressure exerted by N2 gas by van der Waals equation.

(b)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The pressure exerted by N2 gas by van der Waals equation is 187.32atm .

Explanation of Solution

Given

The mass of nitrogen gas is 120kg(=120×103g) .

The number of moles of nitrogen gas is calculated by dividing the mass of nitrogen gas by its molar mass.

    MolesofN2gas=MassMolarmass

Substitute the value of mass and molar mass of nitrogen gas in the above equation.

    MolesofN2=120× 103g28.0g/mol=4.285×103mol

Pressure exerted by nitrogen gas is calculated by van der Waals equation,

    (P+n2aV2)(Vnb)=nRT

Where,

  • P is the pressure of the gas.
  • n is the number of moles of the gas.
  • V is the volume of gas.
  • R is the gas constant (0.08206Latm/molK) .
  • T is the temperature.
  • a is van der Waals constant (1.39L2atm/mol2) . (refer to table 10.3)
  • b is van der Waals constant (0.0391L/mol) .(refer to table 10.3)
Substitute the values of n,V,T,R,a and b in above equation.

    [( P+ ( 4.285× 10 3 mol ) 2 ×1.39 L 2 atm/mol 2 ( 1100.0L ) 2 )×( 1100.0L4.285× 10 3 mol×0.0391L/mol)]=4.285×103mol×0.08206Latm/molK×553K(P+21.09atm)(933L)=194450L-atmP+21.09atm=194450L-atm933LP=208.41atm21.09atm=187.32atm
Conclusion



The pressure exerted by N2 gas by van der Waals equation is 187.32atm .

(c)

Interpretation Introduction

To determine: The dominant correction in the manufacture of ammonia out of the given corrections of finite volume and the attractive interactions.

(c)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The correction of attractive interactions dominates in the manufacture of ammonia.

Explanation of Solution

In the manufacture of ammonia, due to the smaller size and high electronegativity difference of nitrogen and hydrogen the attractive interaction dominates. Nitrogen gas can easily abstract a proton from hydrogen atom due to its high electronegativity. Thereby, hydrogen is highly electropositive, it can easily donate its proton. Therefore, attractive interaction dominated over finite volume correction.

Conclusion

The correction of attractive interactions dominates in the manufacture of ammonia.

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Chapter 19 Solutions

EBK CHEMISTRY:CENTRAL SCIENCE

Ch. 19.5 - Prob. 19.6.1PECh. 19.5 - Prob. 19.6.2PECh. 19.5 - Prob. 19.7.1PECh. 19.5 - Prob. 19.7.2PECh. 19.5 - Prob. 19.8.1PECh. 19.5 - Prob. 19.8.2PECh. 19.6 - Prob. 19.9.1PECh. 19.6 - Prob. 19.9.2PECh. 19.7 - Prob. 19.10.1PECh. 19.7 - Prob. 19.10.2PECh. 19.7 - Prob. 19.11.1PECh. 19.7 - Prob. 19.11.2PECh. 19.7 - Prob. 19.12.1PECh. 19.7 - Prob. 19.12.2PECh. 19 - Prob. 1DECh. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Prob. 3ECh. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - 11.47 Indicate whether each statement is true or...Ch. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10ECh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - A glass vessel fitted with a stopcock valve has a...Ch. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Consider what happens when a sample of the...Ch. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - The normal boiling point of Br2(l) is 58.8 °C, and...Ch. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - 19.35 (a) What do you expect for the sign of ∆S in...Ch. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - For each of the fallowing pairs, choose the...Ch. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Cyclopropane and propylene are isomers that both...Ch. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Using data in Appendix C, calculate HO, SO,...Ch. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - From the values given for H° and S°, calculate G°...Ch. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69ECh. 19 - Prob. 70ECh. 19 - Prob. 71ECh. 19 - Prob. 72ECh. 19 - (a) Use data in Appendix C to estimate the boiling...Ch. 19 - Prob. 74ECh. 19 - Prob. 75ECh. 19 - Prob. 76ECh. 19 - Prob. 77ECh. 19 - Prob. 78ECh. 19 - Consider the reaction 2 NO2(g) (N2O4(g). (a)...Ch. 19 - Prob. 80ECh. 19 - Prob. 81ECh. 19 - Prob. 82ECh. 19 - Prob. 83ECh. 19 - Prob. 84ECh. 19 - Prob. 85ECh. 19 - The Kb, for methylamine (CH3NH2) at 25oC is given...Ch. 19 - Prob. 87AECh. 19 - Prob. 88AECh. 19 - Prob. 89AECh. 19 - Prob. 90AECh. 19 - Prob. 91AECh. 19 - Prob. 92AECh. 19 - Prob. 93AECh. 19 - Prob. 94AECh. 19 - Prob. 95AECh. 19 - Prob. 96AECh. 19 - Consider the following three reactions: (i) Ti (s)...Ch. 19 - Prob. 98AECh. 19 - (a) For each of the following reactions, predict...Ch. 19 - Prob. 100AECh. 19 - Prob. 101AECh. 19 - Prob. 102AECh. 19 - Prob. 103AECh. 19 - Prob. 104AECh. 19 - Prob. 105AECh. 19 - Prob. 106AECh. 19 - Most liquids follow Trouton’s rule (see Exercise...Ch. 19 - 19.108 In chemical kinetics the entropy of...Ch. 19 - Prob. 109IECh. 19 - Prob. 110IECh. 19 - Prob. 111IECh. 19 - Prob. 112IECh. 19 - The following data compare the standard enthalpies...Ch. 19 - Prob. 114IECh. 19 - Prob. 115IECh. 19 - Prob. 116IE
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