Chapter 03 the First Law of Thermodynamics (Pp 59-81)

CH 03 THE FIRST LAW OF THERMODYNAMICS 59

CHAPTER 03 THE FIRST LAW OF THERMODYNAMICS

3-1 THE FIRST LAW OF THERMODYNAMICS

Problem 3-1

A system receives calories200 of heat and work done by the

system is joules736 . What is change in its internal energy?

)186.41( Jcalorie = . B.U. B.Sc. 2000A

Solution According to first law of thermodynamics

dWdUdQ +=

dWdQdU −=

JdU 2.101736)186.4)(200( =−=

Problem 3-2

A system received joules1254 of heat. Calculate the work

done by the system if the change in its internal energy is

calories200 . B.U. B.Sc. 2007A


dWdUdQ +=

dUdQdW −=

Now JdQ 1254=

JJcaloriesdU 2.837)186.4)(200(200 ===

Hence

JdW 8.4162.8371254 =−=

CH 03 THE FIRST LAW OF THERMODYNAMICS 60 Problem 3-3

A system receives 150 calories of heat and the work done by

the system is 418 joules. What is the change in its internal

energy ( Jcalorie 18.41 = ) B.U. B.Sc. 2009S

Solution According to the first law of thermodynamics

dWdUdQ +=

dWdQdU −=

Now JcaloriesdQ 627)18.4)(150(150 ===

JdU 418=

Therefore

JdU 209418627 =−=

Problem 3-4

An ideal gas expands isothermally, performing J31000.5 ×

of work in the process. Calculate

(a) the change in internal energy of the gas and

(b) the heat absorbed during this expansion.

Solution

(a) 0== INTERNALdEdU because there is no change in

temperature.

(b) dWdUdQ +=

JdQ 33 1000.5)1000.5(0 ×=×+=

Problem 3-5

Let kg00.1 of water be converted to steam by boiling. The

volume changes from an initial value of 331000.1 m−× as a

liquid to 3671.1 m as steam. For this process, find

(a) the work done on the system

(b) the heat added to the system and

(c) the change in internal energy of the system.

(Heat of vapourization 161026.2 −×= kgJ , 1 atmosphere 2310013.1 −× mN ) K.U. B.Sc. 2003

Solution (a) The work done on the system


dVpdW =

JdW 535 10692.1)}1000.1(671.1){10013.1( ×=×−×= −

(b) JLmdQ v

66 1026.2)1026.2)(1( ×=×==

(c) dWdQdU −=

JdU 656 10091.2)10692.1()1026.2( ×=×−×=

Problem 3-6

A W40 heat source is applied to a gas sample for s25 ,

during which time the gas expands and does J750 of work

on its surroundings. By how much does the internal energy

of the gas change?

Solution According to the first law of thermodynamics

dWdUdQ +=

dWdQdU −=

Now

JtPdQ 1000)25)(40( ==∆=

JdW 750=

Hence

JdU 2507501000 =−=


3-2 HEAT CAPACITIES OF AN IDEAL GAS

Example 3-7

In an experiment, mol35.1 of oxygen ( 2O ) are heated at

constant pressure starting at C011 . How much heat must be

added to the gas to double its volume? Given that 114.29 −−= KmolJCV is for oxygen.

Solution According to Charles’s law

==2

2

1

1

T

V

T

Vconstant

11

1

1

1

1

2

2 22

TTV

VT

V

VT =

=

=

The amount of heat added to the system at constant pressure is

given by

)2()( 1112 TTCnTTCndTCnQ PPP −=−==

JTCnQ P

4

1 10127.1)27311)(4.29)(35.1( ×=+==

Example 3-8

Twelve grams of nitrogen )( 2N in a steel tank are heated

from C025 to C

0125 . (a) How many moles of nitrogen are

present? (b) How much heat is transferred to the nitrogen?

(The specific heat of 2N at constant volume is 118.20 −− kgmolJ ).

Solution

(a) 28 grams of 2N is = 1 mol

12 grams of 2N is nmol === 429.028

12

(b) dTCnQ V=

JQ 892)}27325()273125){(8.20)(429.0( =+−+=

CH 03 THE FIRST LAW OF THERMODYNAMICS 63 Example 3-9

The mass of helium atom is kg271066.6 −× . Compute the

specific heat at constant volume for helium gas

(in 11 −− KkgJ ) from the molar heat capacity at constant

volume. (Given that 115.12 −−= KmolJCV ).

Solution

Now 115.12 −−= KmolJCV

The mass of one mole of helium gas is

(mass of helium atom) AN

kg32327 10011.4)10022.6)(1066.6( −− ×=××=

Hence

113

310116.3

10011.4

5.12 −−

−×=

×= KkgJCV

Example 3-10

Propane gas )( 83HC behaves like an ideal gas with 127.1=γ .

Determine the molar heat capacity at constant volume and

the molar heat capacity at constant volume.

Solution We know that

RCC VP +=

But V

P

C

C=γ or VP CC γ= , therefore above expression

becomes

RCC VV +=γ

RCV =− )1(γ

115.651127.1

314.8

1

−−=−

=−

= KmolJR

CVγ

118.73314.85.65 −−=+=+= KmolJRCC VP

CH 03 THE FIRST LAW OF THERMODYNAMICS 64 Example 3-11

The heat capacity at constant volume of a certain

amount of a monoatomic gas is KJ /8.49 .

(a) Find the number of moles of the gas.

(b) What is the internal energy of the gas at KT 300= ?

(c) What is the heat capacity of the gas at constant

pressure?

Solution (a) For monoatomic gas

RnCV2

3=

molesR

Cn V 4

)314.8(5

)8.49(2

3

2≅==

(b) JTCU V

410494.1)300)(8.49( ×===

(c) VVVP CCRnCC3

2+=+= RnCV

2

3=Θ

KJCC VP /83)8.49(3

5

3

5===

Example 3-12

One mole of a monoatomic ideal gas is initially at K273

and one atmosphere.

(a) What is its initial internal energy?

(b) Find its final internal energy and the work done by the

gas when J500 of heat are added at constant pressure.

(c) Find the same quantities when J500 of heat are added

at constant volume.

Solution (a) The initial internal energy is given by

TRTCU Vi2

3==

JU i 3405)273)(314.8(2

3==


(b) Now dTCdQ P= or PC

dQdT =

JCC

dQ

C

dQCdTCdU

VPP

VV 300)3/5(

500

)/(===

==

Therefore the final internal energy and work are given by

JdUUU if 37053003405 =+=+=

JdUdQdW 200300500 =−=−=

(c) JdUdQ 500== and 0)0( === pdVpdW


3-2 WORK DONE ON OR BY AN IDEAL GAS 3-2(A) WORK DONE AT CONSTANT VOLUME

Example 3-13

Calculate the increase in internal energy of ten grams of

oxygen whose temperature is increased by C010 at constant

volume. Given that 10182.1 −−= CgJCv .


dWdUdQ +=

dVpdUdTCm v +=

)0()10)(82.1)(10( 0101 pdUCCgJg +=−−

dUJ =182

Increase in internal energy JdUdE INTERNAL 182===


3-2(B) WORK DONE AT CONSTANT PRESSURE

Problem 3-14

A gas expands at atmospheric pressure and its volume

increases by 3500 cm . Calculate the work done by the gas.

(1 Atmosphere25 /10013.1 mN×= ) B.U. B.Sc. 1998A

Solution The work done by the gas at constant pressure is given by

dVpW =

JW 65.50)10500)(10013.1( 65 =××= −

Problem 3-15

One kilogram water is converted to steam at standard

atmospheric pressure. The volume changes from 33101 m

−×

as a liquid to 3671.1 m as steam. For this process calculate

the work done on the system when the pressure is 25 /10013.1 mN× . K.U. B.Sc. 1998

Solution The work done by the vapourizing water is

)( if VVpdVpW −==

JW 535 10691.1)101671.1)(10013.1( ×=×−×= −

Problem 3-16

A gas expands at atmospheric pressure and its volume

increases by 3334 cm . Find the work done by the gas. The

atmospheric pressure is 26 /10013.1 cmdynes× .

B.U. B.Sc. (Hons.) 1988A

The work done by the gas, at constant pressure, is given by

)( if VVpdVpW −==

Now

26 /10013.1 cmdynesp ×=

2

22

56

/)10(

10)10013.1(mNp

−

−××= Ndyne 5101 −=Θ


25 /10013.1 mNp ×= 343323 1034.3)10)((334(334 mmcmdV −− ×===

Hence

JdW 834.33)1034.3)(10013.1( 45 =××= −

Problem 3-17

A gas is compressed at a constant pressure of atm8.0 from

litres9 to litres2 . In the process, J400 of energy leaves the

gas by heat.

(a) What is the work done on the gas?

(b) What is the change in its internal energy?

Solution (a) The work done is given by

dVpdW =

JdW 28.567}10)92)}{(10013.1()8.0{( 35 −=×−××= −

(b) The desired change in internal energy of the system is

given by

dUdWdQ +=

dWdQdU −=

JdU 28.167)28.567(400 =−−−=


3-2(C) WORK DONE AT CONSTANT TEMPERATURE

Problem 3-18

A gram molecule of a gas at C077 expands isothermally to

its double volume. Calculate the amount of work done.


Solution The work done is given by

=

i

f

V

VnTRnW λ

JV

VnW 2017

2)27377)(314.8)(1(

0

0 =

+= λ

Problem 3-19

Calculate the work done by an external agent in

compressing moles12.1 of oxygen from volume of

litres4.22 and atm32.1 pressure to litres3.15 at the same

temperature. P.U. B.Sc. 2001

Solution The desired work on the gas by the external agent is given by

=

=

i

f

ii

i

f

V

VnVp

V

VnTRnW λλ

×××= −

4.22

3.15)104.22)}(10013.1(32.1( 35

nW λ

kJJW 142.110142.1 3 −=×−=

Problem 3-20

A sample of gas consisting of moles11.0 is compressed

from a volume of 30.4 m to 30.1 m while its pressure

increases from 10 to )/(40 2mNPa . Calculate the work

done. P.U. B.Sc. 2002, K.U. B.Sc. 2008

Solution

CH 03 THE FIRST LAW OF THERMODYNAMICS 70 It may be noted that

mNTRnVpVp ffii •=== 40

which indicates that the given process is isothermal. The work

done at constant temperature is given by

Jnp

pnTRnW

f

i 5.55)40/10()40( −==

= λλ

Problem 3-21

A balloon contains mol30.0 of helium. It rises, while

maintaining at constant K300 temperature, to an altitude

where its volume has expanded five times. How much work

is done by the gas in the balloon during this isothermal

expansion?

Solution The work done at constant temperature is given by

=

i

f

V

VnTRnW λ

JV

VnW

3

0

0 10203.15

)300)(314.8)(30.0( ×=

= λ

Problem 3-22

One mole of nitrogen gas is compressed isothermally from 2/10 mN to

2/20 mN at C027 . Calculate the work done.

Solution The work done during this compression is given by

−=

i

f

p

pnTRnW λ

JnW 172910

20)27327)(314.8)(1( −=

+−= λ


3-2(D) WORK DONE IN THERMAL ISOLATION OR UNDER ADIABATIC CONDITIONS

Problem 3-23

One mole of oxygen, initially kept at C017 , is adiabatically

compressed so that its pressure becomes ten times.

Calculate

(a) its temperature after the compression and

(b) the work done on the gas.

Given that 111.21 −−= KmolJCv is for oxygen.

Solution (a) For adiabatic process

γγiiff VpVp =

γγγγ )()( iiiiffff VpppVppp−− =

γγγγ )()( 11

iiff TRnpTRnp−− =

γγγγiiff TpTp

−− = 11

γγ /)1( −

=

f

i

ifp

pTT

Now KKCTi 290)27317(170 =+==

1.010

1==

f

i

p

p , 40.1=γ and

7

2

40.1

40.111−=

−=

−

γ

γ

Hence KT f 560)1.0)(290( 7/2 == −

(b) dTCmdEdW vINTERNAL ==

JdW 5697)290560)(1.21)(1( =−=

Problem 3-24

Calculate the work done to compress adiabatically one gram

mole of air initially at S.T.P. conditions to half its volume if

40.1=γ for air.

Solution


The initial volume 0V of air at S.T.P. is given by

111 TRnVp =

3

1

11

p

TRnV =

mVV2

501 1024.210013.1

)2730)(314.8)(1( −×=×

+==

The final pressure of air, under adiabatic conditions, is given by

γ

=

f

i

ifV

Vpp

PaV

Vp f

5

40.1

0

05 10673.25.0

)10013.1( ×=

×=

The desired work done is

1−

−=

γ

iiff VpVpW

40.11

)1024.2)(10013.1()}1024.2()5.0){(10673.2( 2525

−

××−×××=

−−

W

JW 1812−=

Problem 3-25

32 m of a gas at 2/100 mN expands according to law

CVP =2.1 where C is a constant, until volume is doubled.

Calculate the work done.

Solution

Now CVP ii =2.1

7.229)2)(100( 2.1 ==C

Hence

7.2292.1 =VP

Vp

7.229=

The work done by the gas is given by


dVVdVpW

V

V

∫ ∫−==

2

1

4

2

2.17.229

[ ] JV

W 43.129242.0

7.229

2.07.229 2.02.0

4

2

2.0

=−

−=

−= −−

−

Problem 3-26

Five litres of argon at C00 are compressed to one litre

adiabatically and reversibly. What will be the final

temperature if 3/5=γ for argon?

Solution For an adiabatic process

γγiiff VpVp =

γγ1i

i

i

f

f

fV

V

TRnV

V

TRn

=

TRnVp =Θ

11 −− = γγiiff VTVT

1−

=

γ

f

i

ifV

VTT

KT f 798)1/5)(273( 1)3/5( == −

Problem 3-27

Calculate the final temperature of a sample of carbon

dioxide of mass g0.16 that is expanded reversibly and

adiabatically from 3500 cm at K15.298 to

32000 cm .


1−

=

γ

f

i

ifV

VTT

KT f 71.1962000

500)15.298(

130.1

=

=

−

Problem 3-28


By how much must the volume of a gas with 40.1=γ be

changed to an adiabatic process if the Kelvin temperature is

to double?


γγiiff VpVp =

γγ1i

i

i

f

f

fV

V

TRnV

V

TRn

=

TRnVp =Θ


f

i

i

f

T

T

V

V=

−1γ

)1/(1 −

=

γ

f

i

i

f

T

T

V

V

177.02

)140.1/(1

=

=

−

T

T

V

V

i

f

if VV 177.0=

Problem 3-29

A mol00.1 sample of an ideal diatomic gas originally at

atm00.1 and C020 , expands adiabatically to twice its

volume. What are final pressure and temperature for the

gas? Assume no molecular vibrations.


γγiiff VpVp =

γ

=

f

i

ifV

Vpp


40.1

0

0

2)1(

=

V

Vp f 40.1=γΘ for a diatomic gas.

atmp f 379.0=

Further


140.1

0

0

1

2)27320(

−−

+=

=

V

V

V

VTT

f

i

if

γ

KT f 222=

Problem 3-30

An ideal gas initially at atm00.8 and K300 is permitted to

expand adiabatically until its volume doubles. Find the final

pressure and temperature if the gas is

(a) Monoatomic (b) Diatomic

Solution The expressions for final pressure and temperature of the gas

are given by γ

=

f

i

ifV

Vpp

and

1−

=

γ

f

i

ifV

VTT

(a) For monoatomic gas 3

5=γ therefore

atmV

Vp

i

i

f 52.22

)00.8(

3/5

=

=

KV

VT

i

if 189

2)300(

1)3/5(

=

=

−


(b) For diatomic gas 40.15

7==γ therefore

atmV

Vp

i

i

f 03.32

)00.8(

40.1

=

=

KV

VT

i

i

f 2272

)300(

140.1

=

=

−

Problem 3-31

A volume of dry air at S.T.P. is expanded to three times its

original volume under adiabatic conditions. Calculate the

final temperature and pressure if 40.1=γ for air.

Solution

The final temperature of air is given by

1−

=

γ

f

i

ifV

VTT

KV

VT f 176

3)2730(

40.1

0

0 =

+=

The final pressure of the gas is given by

γ

=

f

i

ifV

Vpp

PaV

Vp f

4

40.1

0

05 10176.23

)10013.1( ×=

×=

Problem 3-32

An ideal monoatomic gas for which 3/5=γ undergoes an

adiabatic expansion to one third of its initial pressure. Find

the ratio of final volume to initial volume if the process is

(a) Isothermal

(b) Adiabatic. K.U. B.Sc. 2000

CH 03 THE FIRST LAW OF THERMODYNAMICS 77 Solution

(a) For an isothermal process

iiff VpVp =

3)3/1(

===i

i

f

i

i

f

p

p

p

p

V

V

(b) For an adiabatic process

γγiiff VpVp =

f

i

i

f

p

p

V

V=

γ

933.1)3()3/1(

6.0

)3/5/(1/1

==

=

=

i

i

f

i

i

f

p

p

p

p

V

Vγ

Example 3-33

A volume of argon gas at C027 expands adiabatically until

its volume is increased four times. Find the resulting fall in

temperature. Given that 67.1=γ is for argon gas.

Solution

Now

1−

=

γ

f

i

ifV

VTT

CorKV

VT f

0

167.1

0

0 5.1545.1184

)27327( −=

+=

−

Example 3-34

An ideal gas at K300 is compressed adiabatically to half

its initial volume.

(a) What is the final temperature of the gas if it is

monoatomic?

(b) What is the final temperature of the gas if it is diatomic?

CH 03 THE FIRST LAW OF THERMODYNAMICS 78 Solution

Now

1−

=

γ

f

i

ifV

VTT

(a) For monoatomic gas 3

5=γ therefore

KV

VT f 476

5.0)300(

1)3/5(

0

0 =

=

−

(b) For diatomic gas 40.15

7==γ therefore

KV

VT f 396

5.0)300(

140.1

0

0 =

=

−

Example 3-35

Calculate the rise in temperature when a gas at C027 is

compressed to eight times its original pressure. The value of

γ is 1.5 for the given gas.

Solution The final temperature of the gas is given by the expression

γγ /)1( −

=

f

i

ifp

pTT

Kp

pT

i

i

f 6008

)27327(

5.1/)5.11(

=

+=

−

The rise in temperature of the gas will be

KTTT if 300300600 =−=−=∆

Example 3-36

A given mass of gas at C00 is suddenly compressed to a

pressure twenty times the initial pressure. What will be the

final temperature of the gas if γ is 42.1 .

CH 03 THE FIRST LAW OF THERMODYNAMICS 79 Solution The final temperature of the gas is given by the expression

γγ /)1( −

=

f

i

ifp

pTT

Kp

pT

i

i

f 2.66220

)2730(

42.1/)42.11(

=

+=

−

Example 3-37

One mol of an ideal monoatomic gas at K300 and atm0.3

expands adiabatically to a final pressure of atm0.1 . How

much work does the gas do in the expansion?

Solution For adiabatic process

γγ /)1( −

=

f

i

ifp

pTT

KT f 1930.1

0.3)300(

67.1/)67.11(

=

=

−

The work done in adiabatic process is given by

JTTR

W fi 1328167.1

)193300(314.8)(

1=

−

−=−

−=

γ

Example 3-38

An ideal monoatomic gas, consisting of mol6.2 of volume 3084.0 m , expands adiabatically. The initial and final

temperatures are C025 and C

068− respectively. What is

the final volume of the gas?

Solution For adiabatic process



11 −−

= γγ

i

f

i

f VT

TV

)1/(1 −

=

γ

f

i

fT

TV

Now

3084.0 mVi =

KKCTi 298)27325(250 =+==

KKCT f 205)27368(680 =+−=−=

67.1=γ for monoatomic gas.

Hence

3

)167.1/(1

2 147.0)084.0(205

298mV =

=

−

CH 03 THE FIRST LAW OF THERMODYNAMICS 81 ADDITIONAL PROBLEMS

(1) A system receives 150 calories of heat and the change

in its internal energy is 209 joules. Calculate the

work done by the system. ( Jcalorie 18.41 = )

B.U. B.Sc. 2007A

(2) The atmospheric pressure is 26 /10013.1 cmdynes× . A

gas expands at this pressure and the increase in its

volume is ..668 cc Find the work done by the gas.


(3) A gas expands at atmospheric pressure and its

volume increases by3400 cm . Find the work done by

the gas. B.U. B.Sc. (Hons.) 1991A

(4) A gas is suddenly compressed to one fourth of its

original volume. Calculate rise in temperature, the

original at C027 and 5.1=γ . F.P.S.C. 1978

Answers

(1) 418 J (2) 67.67 J (3) 40.52 J

(4) 600 K

Documents

Chapter 03 the First Law of Thermodynamics (Pp 59-81)