Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 5, Problem 5.29QP

(a)

Interpretation Introduction

Interpretation: The PV work done by the gas against an external pressure of 1.0atm and the change in internal energy (ΔE) for the system are to be calculated.

Concept introduction: Internal energy of a system is defined as the sum of all the kinetic energy and the potential energy of all the compounds of the system. From the first law of thermodynamics, the change in internal energy is the sum of heat absorbed by the system and the work done on the system.

To determine: The PV work done by N2 gas against an external pressure of 1.0atm .

(a)

Expert Solution
Check Mark

Answer to Problem 5.29QP

Solution

The PV work during the reaction is 126.45J_ .

Explanation of Solution

Explanation

Given

Mass of NaN3 is 2.25g .

External pressure is 1.0atm .

Density of nitrogen is 1.165g/L .

Temperature of the reaction is 20οC .

The conversion of Celsius to Kelvin is done as,

0οC=273K

Therefore, the conversion of 20οC to Kelvin is done as,

20οC=20+273K=293K

The given chemical reaction is,

2NaN3(s)2Na(s)+3N2(g)

The number of moles of NaN3 is calculated by using the formula,

NumberofmolesofNaN3=MassMolarmass

Molar mass of NaN3 is 65g/mol .

Substitute the values of mass and molar mass of NaN3 in the above formula to calculate its number of moles.

NumberofmolesofNaN3=2.25g65g/mol=0.03462mol

As per the balanced chemical reaction, two moles of NaN3 produces three moles of N2 . Therefore, number of moles of N2 formed for 0.03462mol of NaN3 is,

NumberofmolesofN2=0.03462molNaN3×3molN22molNaN3=0.03462×32mol=0.103862=0.05193mol

Therefore, grams of N2 formed is calculated by using the formula,

Mass=Molarmass×Numberofmoles

Substitute the values of molar mass and number of moles in the above formula to calculate the mass of N2 .

MassofN2=28g/mol×0.05193mol=1.45404g

The volume of N2 generated is calculated by using the formula,

Volume=MassDensity

Substitute the value of mass and density of N2 in the above formula to calculate the volume of N2 .

Volume=1.45404g1.165g/L=1.248L

The PV work done during the reaction is calculated by using the expression,

w=PΔV

Where,

  • w is the work done.
  • P is the pressure applied on the system.
  • ΔV is the change in volume of the system.

The change in volume, ΔV=VfVin .

Where,

  • Vf is the final volume.
  • Vin is the initial volume.

Here, initial volume is zero.

Substitute the values of P and ΔV in the above expression to calculate the work done.

w=1.0atm×(VfVin)=1.0atm×(1.248L0)=1×1.248atm.L=1.248atm.L

The conversion of atm.L to joule is done as,

1atm.L=101.33J

Therefore, the conversion of 1.248atm.L to joule is done as,

1.248atm.L=1.248×101.33J=126.45J_

Hence, the PV work during the reaction is 126.45J_ .

(b)

Interpretation Introduction

To determine: The change in internal energy (ΔE) for the system.

(b)

Expert Solution
Check Mark

Answer to Problem 5.29QP

Solution

The change in internal energy, ΔE is 2.466kJ_ .

Explanation of Solution

Explanation

Given

The heat released is 2.34kJ .

The work done is 126.45J .

The conversion of J to kJ is done as,

1J=103kJ

Therefore, the conversion of 126.45J to kJ is done as,

126.45J=126.45×103kJ=0.126kJ

The change in internal energy is calculated by using the first law of thermodynamics which is represented by the equation,

ΔE=qw

Where,

  • ΔE is the change in internal energy.
  • q is the heat released by the system.
  • w is the work done.

Here, heat is released so q is negative.

Substitute the values of q and w in the above equation to calculate the change in internal energy.

ΔE=2.34kJ0.126kJ=2.466kJ_

Hence, the change in internal energy, ΔE is 2.466kJ_ .

Conclusion

  1. a. The PV work during the reaction is 126.45J_ .
  2. b. The change in internal energy, ΔE is 2.466kJ_

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Chapter 5 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 5.6 - Prob. 11PECh. 5.7 - Prob. 12PECh. 5.7 - Prob. 13PECh. 5.8 - Prob. 14PECh. 5.8 - Prob. 15PECh. 5.9 - Prob. 16PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.63QPCh. 5 - Prob. 5.64QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143APCh. 5 - Prob. 5.144AP
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