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Ideal gas of Diatomic Molecules - Physics for Scientist and Engineers - Solved Past Paper, Exams of Engineering Physics

This is the Solved Past Paper of Physics for Scientist and Engineers which includes Ideal Gas of Diatomic Molecules, Avogadro Number, Boltzmann Constant, Universal Gas Constant, Thermal Linear Expansion etc. Key important points are: Ideal Gas of Diatomic Molecules, Avogadro Number, Boltzmann Constant, Universal Gas Constant, Thermal Linear Expansion, Diatomic Molecules, Constant Volume

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Physics 7b
Fall 2003
Midterm 1
Section 1
September 29, 6:00-8:00pm
Answer all problems. Write neatly and clearly, explaining your work.
Partial credit for incomplete solutions will be granted provided your logic
is reasonable and clear. Indicate any parts what should not be counted
toward your grade with an “X”.
Enclose all answers in boxes. All numerical answers must be expressed
in SI units. Answers with no explanations, or disconnected comments
will not receive credit. If you obtain an answer that is questionable,
explain why you think it is wrong.
NAME:__________________________________________________
SID#:___________________________________________________
1:________________/5 8:________________/10
2:________________/5 9:________________/25
3:________________/10 10:_______________/25
4:________________/5
5:________________/5 TOTAL SCORE:___________
6:________________/5
7:________________/5
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Download Ideal gas of Diatomic Molecules - Physics for Scientist and Engineers - Solved Past Paper and more Exams Engineering Physics in PDF only on Docsity!

Physics 7b

Fall 2003

Midterm 1

Section 1

September 29, 6:00-8:00pm

Answer all problems. Write neatly and clearly, explaining your work.

Partial credit for incomplete solutions will be granted provided your logic

is reasonable and clear. Indicate any parts what should not be counted

toward your grade with an “X”.

Enclose all answers in boxes. All numerical answers must be expressed

in SI units. Answers with no explanations, or disconnected comments

will not receive credit. If you obtain an answer that is questionable,

explain why you think it is wrong.

NAME:__________________________________________________

SID#:___________________________________________________

1:________________/5 8:________________/

2:________________/5 9:________________/

3:________________/10 10:_______________/

4:________________/

5:________________/5 TOTAL SCORE:___________

6:________________/

7:________________/

CONSTANTS

Avogadro’s Number: NA= 6.022 × 10

23 / mole

Boltzmann’s Constant: k = 1.381 × 10

  • J/ K = 1.99cal / K*mol

Stefan-Boltzmann’s Constant:

8 2 4 σ 5.67*10 Wm K

− − −

Universal Gas Constant: R = 8.314 J / mol·K = 8.206 × 10

  • l·atm / mol·K

Gravity: g = 9.8 m/s

2

1 cal=4.18J

1 atm=1.013*

5 N-m

1 liter = 10

  • m

3

Material properties

cwater = 4190 J/kg-K kcopper = 2.0 W/m-C

o kair = 0.023 W/m-C

o

kwater = 0.56 W/m-C

o Lvapor(water)=2260 J/kg Atomic Weights

: N 2 :28gm/mol He:4gm/mol

FORMULAS

Thermal Linear Expansion: ∆ L =αL∆T

Work: W = pdV

Specific heat for solids: Q=mc∆T

Maxwell Distribution

Molar specific heat for gases

Q=nC∆T, where C= CV for constant volume, C=CP for constant pressure

Adiabatic expansion: =constant

Heat Transfer

Stefan’s Law:

4 P = σεAT

Thermal conductivity:

dQ dT kA dt dx

= −

No advanced calculator features may be used at any time.

3/2^2 (^2 )

8

4 2

mv v (^) kT

kT v m

dN m N v e dv kT

π

π π

=

  = (^)    

  1. (10pt)You place a 1.5 kg stone in a large lake of water at a temperature of 20°C.

The heat capacity of the stone is 760 J/kg-K and its initial temperature is 5°C.

When this system reaches equilibrium how much has the entropy of the water

changed?

( )

(1.5 )(760 / )(20 5 )

17,100 ( )

17100 58.96 /

293

stone stone stone fstone istone

o

stone reservoir

reservoir

Q m c T T

kg J kg K C

J heat flowed in

Q J S J K

T K

= −

= − −

= +

−∆ − ∆ = = = −

  1. (5 pt) A raindrop falls vertically a distance of 9000 m. Assuming that the change

in potential energy goes entirely into internal energy of the droplet, determine the

change in temperature of the droplet due to this descent.

2

: 0

(9.8 / )(9000 ) 21.

4190 J/kg-K

water

water

water

GIVEN W

PE mgh

U Q mc T

gh c T

gh m s m T K

c

=

∆ =

∆ = = ∆

= ∆

⇒ ∆ = = =

  1. (5 pt) An aluminum cup (linear expansion coefficient=25x

    /°C), capable of 

holding a fluid volume of 100 cm

3 , is filled to the brim with a fluid whose volume

expansion coefficient is 4x

  • /°C. How much, if any, fluid will spill out of the

cup when the temperature is raised from 300 K to 330 K?

5

4

3

4 3

3

3 7.510 /*

410 /*

100

(4 0.75)10 / 30 100*

0.

o Al Al

o

fluid

ifluid iAl

ffluid fAl

fluid Al

o

C

C

V T V V V cm

V

Amount spilled V V

T V

C K cm

cm

β α

β

β

β β

= =

=

∆ = ∆ = = =

= ∆ − ∆

= − ∆

= ^ −   

=

g

  1. (5pt) The ancient Egyptians hung grass mats in doorways and windows and kept them wet to cool the dry summertime air coming into their homes. Explain the

method used here.

The principle here is evaporative cooling, where the

cooling “medium” is the mat. The hot air flowing

through the mat has its temperature lowered because

heat from the “medium” is used to evaporate the water

and cool the mat. The rate at which water is absorbed

into the air passing through the cooling media is the

evaporation rate. Can be used here because have dry

air, so low relative humidity in environment. We are

able to add vapor to the air without coming close to the

saturation pressure.

  1. (10pt) A solid cylindrical copper rod 10cm long has one end maintained at a

temperature of 20K. The other end is blackened and exposed to thermal radiation

at 400K. Assume no heat is radiated or conducted out of the curved surface of the cylinder. When equlibrium is reached, what is the temperature difference between

the two ends?

( ) ( )

4

8 2 4

2

1 1

20 -1 1

5

rad rad

rad

conduct rad rad end

conduct rad

rad end

dQ

P AT

dt

Wm K

for black T K

dQ

Wm A

dt

dQ kA WK cm

T T A T K

dt l m

dQ dQ

dt dt

K K T

− − −

− −

− −

Note: k(copper)=100W/Kcm

1 1

1

1 1 1

: ,

:

: ,


b a o

a a a

b b b

a o b a b

b

o c c a a o o c

c

ISOTHERMAL T constant along isotherm

T T T

Ideal gas Law PV nRT

P V nRT

P V nRT

V P P P P

V B

Adiabatic PV

regrouping PV V nRT V by Ideal gas Law

TV constant

T

T V T V T V V

T

γ

γ γ

γ

γ γ γ

− −

− − −

=

⇒ = =

=

=

= ⇒ =

=

⇒ =

 

= = ⇒ =  

 

1

1

3 / 2

3 / 2 1

5 / 2 5 / 2

2 2

1 1 1

: 3

:

o

p p v

v v

o c

c o

c c c c

o o o c o c

o

d^ d

V

C C C (^) R

C C R

T

Given T Also d for monatomic gas by equipartition

A

so V A V

To calculate P use IGL P V nRT

T nRT P P nR A V P

A A V A

γ

γ

− −

− = − = = =

= =

=

=

= ⇒ = =

b) (10 pt) Complete the following table, expressing heat, work, and internal energy in

terms of the factors A and B, Po and Vo only. Include full and complete calculations for

each entry based on 1

st and 2

nd Law formulae, and process properties.

Process ∆U Q

a-b (^0) PoVoln(B)

b-c

3 1 1

2

o o

P V

A

  −     5 / 2

o o

o o

P V

A

P V

B A

c-a

3 1 1

2

o o

P V

A

  − −    

Process W ∆S

a-b PoVoln(B) nRln(B)

b-c

5 / 2

3 1

2 2

o o

P V

B A

  − −    

-nRln(B)

c-a 3 1 1

2

o o

P V

A

  −    

:

:

2

, 0.

0

ln

ln( ) ln( )

:

0, ln( )

a b

a b

a b

b b b a b a a a a a

a b o a b o o

a b a b

a b o o

ISOTHERMAL

a b

d U nR T by equipartition

For an isothermal process T

U

dV V

W PdV nRT nRT

V V

W nRT B W P V B

FLT U Q W

But U so Q P V B

Entropy calc

− −

− −

∆ = ∆

∆ =

⇒ ∆ =

 

= = = (^)  

 

= ⇒ =

∆ = −

∆ = =

∫ ∫

( )

( )

( , !!) :

:

2

ln ln

2

,

ln

a b

a b

a b

a b

a b

a b a b

a b a b

b b b

a a a

b a o

b

a

ulation

full derivation not required just understanding

FLT dQ dU dW

dQ (^) dU dW d dT dV

nR nR

T T T T V

dQ (^) d T V

S nR nR

T T V

For isothermal process T T T

V S nR

V

− −

− −

= +

  = + = +    

  ∆ = = +    

= =

∆ =

= nR ln( B ) = nR ln( B )

( (^ )^ (^ )(^ ))

( ) ( )

1 2

3 / 2 1 3 / 2 (^2) 5 / 2

b c o

b c o o

c

b c b

b c b b c b c b c

o o o b c o o

o

b c

d U nR T by equipartition

From table in part a

U nRT A

U P V
A

W PdV area under curve

W P V V P P V V
P P P
W V B A V B A
B B A
P V

 ^  

( ) ( )

3 / 2 1 3 / 2 (^2) 5 / 2

5 / 2

5 / 2

o o o

o o

b c b c b c

b c o o o o

B A P V B A
B B A
P V
B A
FLT U Q W
Q P V P V
A B A

− − −

 −^ +^  −^  − 
 ^  
 ^ 

c a

o

c a o o

c a c a o o

a

c a c

c a

Definition o f adiabatic process Q

d U nR T by equipartition

d U nR T

nRT from table in part a A

U P V
A

W U P V by FLT withQ

A

Also W PdV

Can write adiabat

− −

( )

1 1 1 1

i i

a

c a (^) c i i

c c a c c c a c c c c a

c a c c a a o o

b c a c a (^) a

equationas PV PV

dV W PV

V

P V V V P V V P V V

W from adiabat eqn

d W P V P V P V A

Entropy calculation no need to repeat derivation

dQ d S nR T

γ γ

γ γ

γ γ γ γ γ γ γ

γ γ

− − − −

− −

∫ (^ )

( )

( )

0 3 / 2

3 / 2

ln ln

ln ln 2 ( / )

ln( ) ln 0 2

a a

c c

o c a o o

c a

T V

nR T V

For adiabatic process

d T V S nR nR T A A V

nR A nR A

Could also just say S

for adiabatic process by definition since Q

c)

(10 pt) Calculate the efficiency of the engine operating on this cycle. Is it a Carnot engine

and why?

5 / 2

5 / 2

ln( )

ln( )

L

H H

L o o o o

H o o

L

H

W Q

e

Q Q

Q PV P V

A B A

Q P V B

B

e

A B A

For a Carnot engine

T

e

T A

This clearly is not a Carnot engine

Why Have a non adiabatic isothermal porti

L

L

on

Efficiency is lower than a Carnot because now have

Q exhausted at varying T on portion b c rather

than to single T Reversible engines are defined in course

as represented on P V diagram the engine is reversible

b) (5 pt) What is the ratio of vrms for molecules in Boxes A and B before thermal

contact? After thermal contact?

Important that they mention how vrms derived.

  • Internal energy is wholly kinetic energy

for gases.

  • Translational kinetic energy=(1/2)mv
  • By equipartition, have translational

KE=(3/2)kT

2 2 2 2 2 2 2

3

,

1

4 / 28

,

(3/7) 4 / 28 (3/49)

rms

N He

N

rms

He (^) N

rms

He

N He

N

N (^) He

rms

He N rms

He

kT v

m

After thermal equilibrium T T

v

v (^) m

m

BEFORE thermal equilibrium T T

T

v (^) T

v m

m

=

=

= =

= = =

c) (5 pt) How much heat entered or left the nitrogen? The helium?

There’s no contact with the universe, AND no work

done. So change in internal energy=heat.

Heat is added to nitrogen, removed from helium, which

is calculated in part a

( )

A f o o

B f o o

Q R T T RT

Q R T T RT

d) (8 pt)What is the entropy change of the system and the entropy change of the universe?

In this problem, which is thermally isolated, there

is NO difference between the universe and the

system.

ln ln

ln ln

o A o o B o

universe A B

T

S R R

T

T

S R R

T

S S S R

  ^ 

  ^ 