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Question 16: Suppose 60.0ghydrogenbromide, HBr(g), is heated reversibly from 300Kto 500Kat a constant volume of50.0L , and then allowed to expand isothermally and reversibly until the original pressure is reached. Using cp(HBr(g))=29.1JK-1mol-1, calculate ΔU,q,w,ΔH, and ΔSfor this process. Assume that is an ideal gas under these conditions.

Short Answer

Expert verified

The value of change in internal energy, heat, work and change in enthalpy are given below.
ΔU=3084.64Jq=3084.64Jw=-3084.64JΔH=0
The value of change in entropy isS=3.1579JK

Step by step solution

01

Given information

The value of mass of hydrogen bromide is60.0g

The value of volume, temperature is changes from 300Kto500K .

02

Concept of reversible process

A reversible process in thermodynamics is a process that involves a system and its surroundings and whose direction can be changed by minute adjustments to certain environmental factors, such as pressure or temperature.

03

Calculate change in internal energy

The change in internal energy is

The formula used is,
cv=cp-R\hfillΔT=T2-T1\hfilln=numberofmoles
Calculation
ΔU=(0.742mol)×cp-R×T2-T1=(0.742mol)×29.1JK-1mol-1-8.314JK-1mol-1×(500K-300K)=3084.64J

04

Calculate heat

Constant volume process,

q=3084.64J

05

Calculate work done

first law of thermodynamics

U=q+w

isothermal process,ΔU=0,w=-q

Since work done is equal to heat evolved

w=-q=-3084.64J

06

Calculate change in enthalpy

Since, change in internal energy is equal to change in enthalpy

So,ΔU=ΔH=3084.64J

07

Calculate change in entropy

For calculation of change in entropy first calculate initial pressure and final pressure P2.
P1V=nFT1P1=nRT1VP1=0.742mol×8.314JK-1mol-1×300K50.0L=0.365atm
Final pressure is,
P2V=nRT2P2=nRT2VP2=0.742mol×8.314JK-1mol-1×500K50.0L=0.609atm
The change in entropy is given as ΔS=nRlnV2V1

ΔS=0.742mol×8.314JK-1mol-1ln0.6090.365=0.742mol×8.314JK-1mol-1ln0.6090.365=3.1579JK



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Most popular questions from this chapter

In Example 12.3, a process was considered in which 72.4 g iron initially at 100.0°C was added to 100.0 g water initially at 10.0°C, and an equilibrium temperature of 16.5°C was reached. Take cpFeto be 25.1JK-1mol-1and cp(H2O)to be 75.3JK-1mol-1, independent of temperature. Calculate for the iron, for the water, andStotal in this process.

Question: The molar enthalpy of fusion of ice at 0°Cis6.02kJmol1; the molar heat capacity of under cooled water is75.3Jmol1K1.

(a) One mole of under cooled water at10°Cis induced to crystallize in a heat-insulated vessel. The result is a mixture of ice and water at0°C. What fraction of this mixture is ice?

(b) Calculate ΔSfor the system.

(a) Use data from Appendix D to calculate H°andS°at25°Cfor the reaction

2CuCl2(s)2CuCl(s)+Cl2(g)

(b)Calculate atG°590K, assumingH°andS°are independent of temperature.

(c) Careful high-temperature measurements show that when this reaction is performed at590K,H590°is158.36kJandS°590is177.74JK-1. Use these facts to compute an improved value ofG°590for this reaction. Determine the percentage error inG°590that comes from using the298Kvalues in place of590-Kvalues in this case.

At 12000C, the reduction of iron oxide to elemental iron and oxygen is not spontaneous:

2Fe2O3(s)4Fe(s)+3O2(g)ΔG=+840kJ

Show how this process can be made to proceed if all the oxygen generated reacts with carbon:

C(s)+O2(g)CO2(g)ΔG=-400kJ

This observation is the basis for the smelting of iron ore with coke to extract metallic iron.

The normal boiling point of liquid ammonia is240K ; the enthalpy of vaporization at that temperature is23.4kJmol-1 . The heat capacity of gaseous ammonia at constant pressure is38Jmol-1K-1 .

(a) Calculateq,w,ΔH , and Ufor the following change in state:

2.00molNH3(,1atm,240K)2.00molNH3(g,1atm,298K)

Assume that the gas behaves ideally and that the volume occupied by the liquid is negligible.

(b) Calculate the entropy of vaporization of NH3at240K .

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