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CHAPTER 03

THE FIRST LAW OF THERMODYNAMICS

3-1 THE FIRST LAW OF THERMODYNAMICS Problem 3-1

A system receives 200 caloriesof heat and work done by the system is736 joules. What is change in its internal energy?

) 186 . 4 1

( calorie= J . B.U. B.Sc. 2000A

Solution

According to first law of thermodynamics dQ=dU+dW

dU =dQdW

J

dU =(200)(4.186)−736=101.2

Problem 3-2

A system received 1254 joulesof heat. Calculate the work done by the system if the change in its internal energy is

calories

200 . B.U. B.Sc. 2007A Solution

According to first law of thermodynamics dQ=dU+dW dW =dQdU Now dQ=1254 J J J calories dU =200 =(200)(4.186) =837.2 Hence dW =1254−837.2=416.8 J

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Problem 3-3

A system receives 150calories of heat and the work done by the system is 418joules. What is the change in its internal energy (1 calorie=4.18 J) B.U. B.Sc. 2009S Solution

According to the first law of thermodynamics dW dU dQ= + dW dQ dU = − Now dQ=150 calories=(150)(4.18)=627 J dU =418 J Therefore J dU =627−418=209 Problem 3-4

An ideal gas expands isothermally, performing 5.00×103J 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) dU =dEINTERNAL =0 because there is no change in temperature.

(b) dQ=dU +dW

J dQ=0+(5.00×103)=5.00×103 Problem 3-5

Let 1.00 kg of water be converted to steam by boiling. The volume changes from an initial value of 1.00×10−3m3as a liquid to 1.671 m3as 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 =2.26×106J kg−1, 1 atmosphere

2 3 10 013 . 1 × N m) K.U. B.Sc. 2003 Solution

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dW = p dV J dW =(1.013×105){1.671−(1.00×10−3)}=1.692×105 (b) dQ=m Lv =(1)(2.26×106)=2.26×106J (c) dU =dQdW dU =(2.26×106)−(1.692×105)=2.091×106J Problem 3-6

A 40 W heat source is applied to a gas sample for25 s, during which time the gas expands and does 750 J of work on its surroundings. By how much does the internal energy of the gas change?

Solution

According to the first law of thermodynamics dQ=dU+dW dU =dQdW Now dQ=Pt =(40)(25)=1000 J dW =750 J Hence dU =1000−750=250 J

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3-2 HEAT CAPACITIES OF AN IDEAL GAS Example 3-7

In an experiment, 1.35 molof oxygen (O ) are heated at 2

constant pressure starting at110C. How much heat must be added to the gas to double its volume? Given that

1 1 4 . 29 − − = J mol K CV is for oxygen. Solution

According to Charles’s law = = 2 2 1 1 T V T V constant 1 1 1 1 1 1 2 2 2 2 T T V V T V V T  =      =       =

The amount of heat added to the system at constant pressure is given by Q=n CP dT =n CP(T2T1)=n CP(2 T1T1) J T C n Q P 4 1 =(1.35)(29.4)(11+273)=1.127×10 = Example 3-8

Twelve grams of nitrogen (N2)in a steel tank are heated from 250C to 1250C. (a) How many moles of nitrogen are present? (b) How much heat is transferred to the nitrogen? (The specific heat of N at constant volume is 2

1 1 8 . 20 J molkg). Solution

(a) 28 grams of N2 is = 1 mol

12 grams of N 2 is = =0.429 mol =n 28 12 (b) Q=n CV dT J Q=(0.429)(20.8){(125+273)−(25+273)}=892

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Example 3-9

The mass of helium atom is 6.66×10−27kg. Compute the specific heat at constant volume for helium gas (inJ kg−1K−1) from the molar heat capacity at constant volume. (Given that CV =12.5 J mol−1K−1).

Solution

Now CV =12.5 J mol−1K−1

The mass of one mole of helium gas is (mass of helium atom) NA

kg 3 23 27 10 011 . 4 ) 10 022 . 6 )( 10 66 . 6 ( × − × = × − = Hence 1 1 3 3 3.116 10 10 011 . 4 5 . 12 − − − = × × = J kg K CV Example 3-10

Propane gas (C3H8)behaves like an ideal gas withγ =1.127. Determine the molar heat capacity at constant volume and the molar heat capacity at constant volume.

Solution We know that R C CP = V + But V P C C =

γ or CPCV , therefore above expression

becomes γ CV =CV +R (γ − )1 CV =R 1 1 5 . 65 1 127 . 1 314 . 8 1 − − = − = − = R J mol K CV γ CP =CV +R=65.5+8.314=73.8 J mol−1K−1

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Example 3-11

The heat capacity at constant volume of a certain amount of a monoatomic gas is 49.8 J/K. (a) Find the number of moles of the gas.

(b) What is the internal energy of the gas at T =300 K? (c) What is the heat capacity of the gas at constant

pressure? Solution

(a) For monoatomic gas CV n R 2 3 = moles R C n V 4 ) 314 . 8 ( 5 ) 8 . 49 ( 2 3 2 ≅ = = (b) U CVT J 4 10 494 . 1 ) 300 )( 8 . 49 ( = × = = (c) CP CV n R CV CV 3 2 + = + = CV n R 2 3 = Θ CP CV (49.8) 83 J/K 3 5 3 5 = = = Example 3-12

One mole of a monoatomic ideal gas is initially at 273 K and one atmosphere.

(a) What is its initial internal energy?

(b) Find its final internal energy and the work done by the gas when 500 Jof heat are added at constant pressure. (c) Find the same quantities when 500 Jof heat are added at constant volume.

Solution

(a) The initial internal energy is given by Ui CVT R T 2 3 = = J Ui (8.314)(273) 3405 2 3 = =

(7)

(b) Now dQ=CPdT or P C dQ dT = J C C dQ C dQ C dT C dU V P P V V 300 ) 3 / 5 ( 500 ) / ( = = =       = =

Therefore the final internal energy and work are given by Uf =Ui +dU =3405+300=3705 J

dW =dQdU =500−300=200 J

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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 100C at constant volume. Given that Cv =1.82 J g−1 0C−1.

Solution

According to first law of thermodynamics dQ=dU+dW m CvdT =dU + p dV ) 0 ( ) 10 )( 82 . 1 )( 10 ( g J g−1 0C−1 0C =dU + p 182 J =dU

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3-2(B) WORK DONE AT CONSTANT PRESSURE Problem 3-14

A gas expands at atmospheric pressure and its volume increases by 500 cm3. Calculate the work done by the gas. (1 Atmosphere 5 2 / 10 013 . 1 × N m = ) B.U. B.Sc. 1998A Solution

The work done by the gas at constant pressure is given by W = p dV

J W =(1.013×105)(500×10−6)=50.65 Problem 3-15

One kilogram water is converted to steam at standard atmospheric pressure. The volume changes from 1×10−3m3 as a liquid to 1.671 m3as steam. For this process calculate the work done on the system when the pressure is

2 5 / 10 013 . 1 × N m . K.U. B.Sc. 1998 Solution

The work done by the vapourizing water is W = p dV = p (VfVi) J W 5 3 5 10 691 . 1 ) 10 1 671 . 1 )( 10 013 . 1 ( × − × = × = − Problem 3-16

A gas expands at atmospheric pressure and its volume increases by334 cm3. Find the work done by the gas. The atmospheric pressure is 1.013×106dynes/cm2.

B.U. B.Sc. (Hons.) 1988A The work done by the gas, at constant pressure, is given by

) (Vf Vi p dV p W = = − Now 6 2 / 10 013 . 1 dynes cm p= × 2 2 2 5 6 / ) 10 ( 10 ) 10 013 . 1 ( m N p − × × = dyne 5N 10 1 = − Θ

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p=1.013×105N/m2 3 4 3 3 2 3 10 34 . 3 ) 10 )(( 334 ( 334 cm m m dV = = − = × − Hence J dW =(1.013×105)(3.34×10−4)=33.834 Problem 3-17

A gas is compressed at a constant pressure of 0.8 atm from litres

9 to2 litres. In the process, 400 Jof 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 dW = p dV dW {(0.8) (1.013 105)}{(2 9) 10 3} 567.28 J − = × − × × = −

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

dQ=dW +dU dU =dQdW

J

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3-2(C) WORK DONE AT CONSTANT TEMPERATURE Problem 3-18

A gram molecule of a gas at 770C expands isothermally to

its double volume. Calculate the amount of work done. B.U. B.Sc. (Hons.) 1987A

Solution

The work done is given by

      = i f V V n T R n W λ J V V n W (1)(8.314)(77 273) 2 2017 0 0 =       + = λ Problem 3-19

Calculate the work done by an external agent in compressing 1.12 moles of oxygen from volume of

litres

4 .

22 and 1.32 atm pressure to 15.3 litresat the same temperature. P.U. B.Sc. 2001 Solution

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

      =       = i f i i i f V V n V p V V n T R n W λ λ       × × × = − 4 . 22 3 . 15 ) 10 4 . 22 )}( 10 013 . 1 ( 32 . 1 ( 5 3 n W λ W =−1.142×103J =−1.142 kJ Problem 3-20

A sample of gas consisting of 0.11 moles is compressed from a volume of 4.0 m3 to 1.0 m3 while its pressure increases from 10 to 40 Pa (N/m2). Calculate the work done. P.U. B.Sc. 2002, K.U. B.Sc. 2008 Solution

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It may be noted that m N T R n V p V pi i = f f = =40 •

which indicates that the given process is isothermal. The work done at constant temperature is given by

J n p p n T R n W f i 5 . 55 ) 40 / 10 ( ) 40 ( =− =         = λ λ Problem 3-21

A balloon contains 0.30 molof helium. It rises, while maintaining at constant 300 Ktemperature, 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 V n T R n W λ J V V n W 3 0 0 10 203 . 1 5 ) 300 )( 314 . 8 )( 30 . 0 ( = ×      = λ Problem 3-22

One mole of nitrogen gas is compressed isothermally from

2 /

10 N m to 2

/

20 N m at 270C. Calculate the work done. Solution

The work done during this compression is given by

      − = i f p p n T R n W λ J n W 1729 10 20 ) 273 27 )( 314 . 8 )( 1 ( =−      + − = λ

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3-2(D) WORK DONE IN THERMAL ISOLATION OR UNDER ADIABATIC CONDITIONS Problem 3-23

One mole of oxygen, initially kept at 170C, 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 Cv =21.1 J mol−1K−1 is for oxygen. Solution

(a) For adiabatic process

pfVfγ = piViγ pf pf−γ (pfVf)γ = pipi−γ (piVi)γ γ γ γ γ ) ( ) ( 1 1 i i f f n R T p n R T p− = − p1f−γ Tfγ = pi1−γ Tiγ γ γ)/ 1 (−         = f i i f p p T T Now Ti =170C =(17+273) K =290 K 0.1 10 1 = = f i p p , γ =1.40 and 7 2 40 . 1 40 . 1 1 1 − = − = − γ γ Hence Tf =(290)(0.1)−2/7 =560 K (b) dW =dEINTERNAL =m CvdT J dW =(1)(21.1)(560−290)=5697 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

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The initial volume V0of air at S.T.P. is given by p1V1 =n R T1 3 1 1 1 p T R n V = m V V1 0 5 2.24 10 2 10 013 . 1 ) 273 0 )( 314 . 8 )( 1 ( = × − × + = =

The final pressure of air, under adiabatic conditions, is given by γ         = f i i f V V p p Pa V V pf 5 40 . 1 0 0 5 10 673 . 2 5 . 0 ) 10 013 . 1 (  = ×      × =

The desired work done is

1 − − = γ i i f fV pV p W 40 . 1 1 ) 10 24 . 2 )( 10 013 . 1 ( )} 10 24 . 2 ( ) 5 . 0 ){( 10 673 . 2 ( 5 2 5 2 − × × − × × × = − − W J W =−1812 Problem 3-25 3

2 m of a gas at 100 N/m2expands according to law C

V P 1.2 =

where C is a constant, until volume is doubled. Calculate the work done.

Solution Now PiVi1.2 =C 7 . 229 ) 2 )( 100 ( 1.2 = = C Hence P V1.2 =229.7 V p=229.7

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W p dV V dV V V

=

− = 2 1 4 2 2 . 1 7 . 229

[

]

J V W 4 2 129.43 2 . 0 7 . 229 2 . 0 7 . 229 0.2 0.2 4 2 2 . 0 = −     − =       − = − − − Problem 3-26

Five litres of argon at 00C are compressed to one litre adiabatically and reversibly. What will be the final temperature if γ =5/3 for argon?

Solution

For an adiabatic process pfVfγ = piViγ γ i1γ i i f f f V V T R n V V T R n       =         Θ p V =n R T TfVfγ−1 =TiViγ−1 1 −         = γ f i i f V V T T K Tf =(273)(5/1)(5/3)−1=798 Problem 3-27

Calculate the final temperature of a sample of carbon dioxide of mass 16.0 gthat is expanded reversibly and adiabatically from 500 cm3at 298.15 Kto2000 cm3.

Solution

For an adiabatic process 1 −         = γ f i i f V V T T K Tf 196.71 2000 500 ) 15 . 298 ( 1 30 . 1 =       = − Problem 3-28

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By how much must the volume of a gas with γ =1.40 be changed to an adiabatic process if the Kelvin temperature is to double?

Solution

For an adiabatic process pfVfγ = piViγ γ i1γ i i f f f V V T R n V V T R n       =         Θ p V =n R T TfVfγ−1 =TiViγ−1 f i i f T T V V =       γ−1 ) 1 /( 1 −         = γ f i i f T T V V 0.177 2 ) 1 40 . 1 /( 1 =       = − T T V V i f Vf =0.177 Vi Problem 3-29

A 1.00 mol sample of an ideal diatomic gas originally at atm

00 .

1 and 200C, expands adiabatically to twice its volume. What are final pressure and temperature for the gas? Assume no molecular vibrations.

Solution

For an adiabatic process pfVfγ = piViγ γ         = f i i f V V p p

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40 . 1 0 0 2 ) 1 (       = V V

pf Θ γ =1.40 for a diatomic gas.

pf =0.379 atm Further TfVfγ−1 =TiViγ−1 1 40 . 1 0 0 1 2 ) 273 20 ( − −       + =         = V V V V T T f i i f γ Tf =222 K Problem 3-30

An ideal gas initially at 8.00 atmand 300 Kis 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 i f V V p p and 1 −         = γ f i i f V V T T

(a) For monoatomic gas

3 5 = γ therefore atm V V p i i f 2.52 2 ) 00 . 8 ( 3 / 5 =       = K V V T i i f 189 2 ) 300 ( 1 ) 3 / 5 ( =       = −

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(b) For diatomic gas 1.40 5 7 = = γ therefore atm V V p i i f 3.03 2 ) 00 . 8 ( 40 . 1 =       = K V V T i i f 227 2 ) 300 ( 1 40 . 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 γ =1.40 for air.

Solution

The final temperature of air is given by 1 −         = γ f i i f V V T T K V V Tf 176 3 ) 273 0 ( 40 . 1 0 0 =       + =

The final pressure of the gas is given by γ         = f i i f V V p p Pa V V pf 4 40 . 1 0 0 5 10 176 . 2 3 ) 10 013 . 1 (  = ×      × = Problem 3-32

An ideal monoatomic gas for which γ =5/3 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

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Solution

(a) For an isothermal process pfVf = piVi 3 ) 3 / 1 ( = = = i i f i i f p p p p V V

(b) For an adiabatic process pfVfγ = piViγ 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 V γ Example 3-33

A volume of argon gas at 270Cexpands adiabatically until its volume is increased four times. Find the resulting fall in temperature. Given thatγ =1.67is for argon gas.

Solution Now 1 −         = γ f i i f V V T T C or K V V Tf 0 1 67 . 1 0 0 118.5 154.5 4 ) 273 27 (  = −      + = − Example 3-34

An ideal gas at 300 K is compressed adiabatically to half its initial volume.

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

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Solution Now 1 −         = γ f i i f V V T T

(a) For monoatomic gas

3 5 = γ therefore K V V Tf 476 5 . 0 ) 300 ( 1 ) 3 / 5 ( 0 0 =       = −

(b) For diatomic gas 1.40 5 7 = = γ therefore K V V Tf 396 5 . 0 ) 300 ( 1 40 . 1 0 0 =       = − Example 3-35

Calculate the rise in temperature when a gas at 270Cis 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 γ γ)/ 1 (−         = f i i f p p T T K p p T i i f 600 8 ) 273 27 ( 5 . 1 / ) 5 . 1 1 ( =       + = −

The rise in temperature of the gas will be

K T T T = fi =600−300=300 ∆ Example 3-36

A given mass of gas at 00Cis suddenly compressed to a pressure twenty times the initial pressure. What will be the final temperature of the gas if γ is 1.42 .

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Solution

The final temperature of the gas is given by the expression γ γ)/ 1 (−         = f i i f p p T T K p p T i i f 662.2 20 ) 273 0 ( 42 . 1 / ) 42 . 1 1 ( =       + = − Example 3-37

One mol of an ideal monoatomic gas at 300 Kand 3.0 atm expands adiabatically to a final pressure of1.0 atm. How much work does the gas do in the expansion?

Solution

For adiabatic process γ γ)/ 1 (−         = f i i f p p T T K Tf 193 0 . 1 0 . 3 ) 300 ( 67 . 1 / ) 67 . 1 1 ( =       = −

The work done in adiabatic process is given by

J T T R W i f 1328 1 67 . 1 ) 193 300 ( 314 . 8 ) ( 1 − = − = − − = γ Example 3-38

An ideal monoatomic gas, consisting of 2.6 molof volume

3 084 .

0 m , expands adiabatically. The initial and final temperatures are 250C and −680Crespectively. What is the final volume of the gas?

Solution

For adiabatic process

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1 1 − −         = γ γ i f i f V T T V ) 1 /( 1 −         = γ f i f T T V Now 3 084 . 0 m Vi = Ti =250C=(25+273)K =298 K K K C Tf =−680 =(−68+273) =205

γ =1.67 for monoatomic gas. Hence 3 ) 1 67 . 1 /( 1 2 (0.084) 0.147 205 298 m V  =      = −

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ADDITIONAL PROBLEMS

(1) A system receives 150calories of heat and the change in its internal energy is 209joules. Calculate the work done by the system. (1 calorie=4.18 J)

B.U. B.Sc. 2007A (2) The atmospheric pressure is 6 2

/ 10

013 .

1 × dynes cm . A

gas expands at this pressure and the increase in its

volume is 668 c.c.Find the work done by the gas. B.U. B.Sc. (Hons.) 1989A

(3) A gas expands at atmospheric pressure and its volume increases by400 cm3. 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 270Cand γ =1.5. F.P.S.C. 1978

Answers

(1) 418 J (2) 67.67 J (3) 40.52 J (4) 600 K

References

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