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### Transcript of Chapter 1: Fundamental Concepts of Thermodynamics Chapter 1: Fundamental Concepts of Thermodynamics

• 1

Chapter 1: Fundamental Concepts of Thermodynamics

Questions on Concepts

Q1.1) The location of the boundary between the system and the surroundings is a choice that

must be made by the thermodynamicist. Consider a beaker of boiling water in an airtight room.

Is the system open or closed if you place the boundary just outside the liquid water? Is the sys-

tem open or closed if you place the boundary just inside the walls of the room?

If the system boundaries are just outside of the liquid water, the system is open because water

can escape from the top surface. The system is closed if the boundary is just inside the walls, be-

cause the room is airtight.

Q1.2) Real walls are never totally adiabatic. Order the following walls in increasing order with

respect to their being diathermal: 1-cm-thick concrete, 1-cm-thick vacuum, 1-cm-thick copper,

1-cm-thick cork.

1-cm-thick vacuum < 1-cm-thick cork < 1-cm-thick concrete < 1-cm-thick copper

Q1.3) Why is the possibility of exchange of matter or energy appropriate to the variable of in-

terest a necessary condition for equilibrium between two systems?

Equilibrium is a dynamic process in which the rates of two opposing processes are equal. How-

ever, if the rate in each direction is zero, no exchange is possible, and the system cannot reach

equilibrium.

Q1.4) At sufficiently high temperatures, the van der Waals equation has the form P ≈ RT/(Vm –

b). Note that the attractive part of the potential has no influence in this expression. Justify this

behavior using the potential energy diagram of Figure 1.7.

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• Chapter 1/Fundemental Concepts of Thermodynamics

2

In this case, the energy well depth is small compared to the total energy of the particle. There-

fore, the particle is unaffected by the attractive part of the potential.

Q1.5) Parameter a in the van der Waals equation is greater for H2O than for He. What does this

say about the form of the potential function in Figure 1.7 for the two gases?

This means that the depth of the attractive potential is greater for H2O than for He.

Problems

Problem numbers in RED indicate that the solution to the problem is given in the Student Solu-

tions Manual.

P1.1) A sealed flask with a capacity of 1.00 dm3 contains 5.00 g of ethane. The flask is so weak

that it will burst if the pressure exceeds 1.00 × 106 Pa. At what temperature will the pressure of

the gas exceed the bursting pressure?

With nRTpV = and M mn = :

r

V(r) E

r

V(r) E

r

V(r) E

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• Chapter 1/Fundemental Concepts of Thermodynamics

3

( ) ( ) ( ) ( ) ( ) K 723.6mol K J 8.314472 kg 0.005

mol kg10 0.083 m 0.001 Pa 101.00 R m M V pT 11

1-336

= ×

××××== −− −

P1.2) Consider a gas mixture in a 2.00-dm3 flask at 27.0°C. For each of the following mixtures,

calculate the partial pressure of each gas, the total pressure, and the composition of the mixture

in mole percent:

a. 1.00 g H2 and 1.00 g O2

b. 1.00 g N2 and 1.00 g O2

c. 1.00 g CH4 and 1.00 g NH3

To calculate the partial pressures we use the ideal gas law:

a)

( ) ( ) ( ) ( ) ( ) Pa 10 6.18mol kg10 2.02 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001 VM

T R m V

T R n p 513-33-

11

2

22

2 ×=

××× ××=== −

−−

H

HH H

( ) ( ) ( ) ( ) ( ) Pa 10 3.90 mol kg10 32.0 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001 VM

T R m V

T R n p 513-33-

11

2

22

2 ×=

××× ××=== −

−−

O

OO O

Pa 10 57.6ppp 5total 22 ×=+= OH

( ) % 4.19

mol kg10 32.0 kg 0.001

mol kg10 2.02 kg 0.001

mol kg10 2.02 kg 0.001

100 O molH mol

H mol100H % mol

13-13-

13-

22

2 2 =

 

  

  

  

 ×

+ 

  

 ×

 

  

 ×

×= +

×=

−−

( ) % .95

mol kg10 32.0 kg 0.001

mol kg10 2.02 kg 0.001

mol kg10 32.0 kg 0.001

100 O molH mol

O mol 100O % mol

13-13-

13-

22

2 2 =

  

   

  

  

 ×

+ 

  

 ×

 

  

 ×

×= +

×=

−−

b)

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• Chapter 1/Fundemental Concepts of Thermodynamics

4

( ) ( ) ( ) ( ) ( ) Pa 10 4.45mol kg10 28.02 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001 VM

T R m V

T R n p 413-33-

11

2

22

2 ×=

××× ××=== −

−−

N

NN N

( ) ( ) ( ) ( ) ( ) Pa 10 3.90 mol kg10 32.0 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001 VM

T R m V

T R n p 513-33-

11

2

22

2 ×=

××× ××=== −

−−

O

OO O

Pa 10 35.8ppp 4total 22 ×=+= OH

( ) % 3.35

mol kg10 32.0 kg 0.001

mol kg10 28.02 kg 0.001

mol kg10 28.02 kg 0.001

100 O molN mol

N mol100N % mol

13-13-

13-

22

2 2 =

 

  

  

  

 ×

+ 

  

 ×

 

  

 ×

×= +

×=

−−

( ) % 6.74

mol kg10 32.0 kg 0.001

mol kg10 28.02 kg 0.001

mol kg10 32.0 kg 0.001

100 O molN mol

O mol 100O % mol

13-13-

13-

22

2 2 =

  

   

  

  

 ×

+ 

  

 ×

 

  

 ×

×= +

×=

−−

c)

( ) ( ) ( ) ( ) ( ) Pa 10 7.32mol kg10 17.03 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001 VM

T R m V

T R n p 413-33-

11

3

33

3 ×=

××× ××=== −

−−

NH

NHNH NH

( ) ( ) ( ) ( ) ( ) Pa 10 7.77 mol kg10 16.04 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001 VM

T R m V

T R n p 413-33-

11

4

44

4 ×=

××× ××=== −

−−

CH

CHCH CH

Pa 10 51.1ppp 5total 22 ×=+= OH

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• Chapter 1/Fundemental Concepts of Thermodynamics

5

( )

% 8.54

mol kg10 17.03 kg 0.001

mol kg10 16.04 kg 0.001

mol kg10 17.03 kg 0.001

100

CH molNH mol NH mol

100NH % mol

13-13-

13-

43

3 3

=

 

  

  

  

 ×

+ 

  

 ×

 

  

 ×

×=

+ ×=

−−

( )

% 1.55

mol kg10 17.03 kg 0.001

mol kg10 16.04 kg 0.001

mol kg10 17.03 kg 0.001

100

CH molNH mol CH mol

100CH % mol

13-13-

13-

43

4 4

=

  

   

  

  

 ×

+ 

  

 ×

 