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 200 calories of heat and work done by the system is 736 joules . What is change in its internal energy? (1 calorie = 4.186 J ) . B.U. B.Sc. 2000A Solution According to first law of thermodynamics dQ = dU + dW dU = dQ − dW dU = (200)(4.186) − 736 = 101.2 J Problem 3-2 A system received 1254 joules of heat. Calculate the work done by the system if the change in its internal energy is 200 calories . B.U. B.Sc. 2007A Solution According to first law of thermodynamics dQ = dU + dW dW = dQ − dU Now

dQ = 1254 J dU = 200 calories = (200)(4.186) J = 837.2 J

Hence dW = 1254 − 837.2 = 416.8 J

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 ( 1 calorie = 4.18 J ) B.U. B.Sc. 2009S Solution According to the first law of thermodynamics dQ = dU + dW dU = dQ − dW dQ = 150 calories = (150)(4.18) = 627 J dU = 418 J Therefore dU = 627 − 418 = 209 J Problem 3-4 An ideal gas expands isothermally, performing 5.00 × 10 3 J 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 = dE INTERNAL = 0 because there is no change in temperature. (b) dQ = dU + dW dQ = 0 + (5.00 × 103 ) = 5.00 × 103 J Problem 3-5 Let 1.00 kg of water be converted to steam by boiling. The Now

volume changes from an initial value of 1.00 × 10 −3 m 3 as a liquid to 1.671 m 3 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 = 2.26 × 10 6 J kg −1 , 1 atmosphere 1.013 × 10 3 N m −2 ) K.U. B.Sc. 2003 Solution (a) The work done on the system

CH 03 THE FIRST LAW OF THERMODYNAMICS 61 dW = p dV dW = (1.013 × 10 5 ){1.671 − (1.00 × 10 −3 )} = 1.692 × 10 5 J (b) dQ = m Lv = (1)(2.26 × 10 6 ) = 2.26 × 10 6 J (c) dU = dQ − dW dU = (2.26 × 10 6 ) − (1.692 × 10 5 ) = 2.091 × 10 6 J Problem 3-6 A 40 W heat source is applied to a gas sample for 25 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 = dQ − dW Now dQ = P∆t = (40)(25) = 1000 J dW = 750 J Hence dU = 1000 − 750 = 250 J

CH 03 THE FIRST LAW OF THERMODYNAMICS 62 3-2 HEAT CAPACITIES OF AN IDEAL GAS Example 3-7 In an experiment, 1.35 mol of oxygen ( O2 ) are heated at constant pressure starting at 110 C . How much heat must be added to the gas to double its volume? Given that CV = 29.4 J mol −1 K −1 is for oxygen. Solution According to Charles’s law V1 V2 = = constant T1 T2  2 V1  V   T1 = 2 T1 T2 =  2  T1 =  V  V1  1   The amount of heat added to the system at constant pressure is given by Q = n C P dT = n C P (T2 − T1 ) = n C P (2 T1 − T1 )

Q = n C P T1 = (1.35)(29.4)(11 + 273) = 1.127 × 10 4 J Example 3-8 Twelve grams of nitrogen ( N 2 ) in a steel tank are heated from 25 0 C to 125 0 C . (a) How many moles of nitrogen are present? (b) How much heat is transferred to the nitrogen? (The specific heat of N 2 at constant volume is 20.8 J mol −1 kg −1 ). Solution (a) 28 grams of N 2 is = 1 mol 12 12 grams of N 2 is = = 0.429 mol = n 28 (b) Q = n CV dT

Q = (0.429)(20.8){(125 + 273) − (25 + 273)} = 892 J

CH 03 THE FIRST LAW OF THERMODYNAMICS 63 Example 3-9 The mass of helium atom is 6.66 × 10 −27 kg . Compute the specific heat at constant volume for helium gas (in J kg −1 K −1 ) from the molar heat capacity at constant volume. (Given that CV = 12.5 J mol −1 K −1 ). Solution Now CV = 12.5 J mol −1 K −1 The mass of one mole of helium gas is (mass of helium atom) N A = (6.66 × 10 −27 )(6.022 × 10 23 ) = 4.011 × 10 −3 kg Hence 12.5 CV = = 3.116 × 10 3 J kg −1 K −1 −3 4.011 × 10 Example 3-10 Propane gas (C 3 H 8 ) behaves like an ideal gas with γ = 1.127 . Determine the molar heat capacity at constant volume and the molar heat capacity at constant volume. Solution We know that C P = CV + R

But γ =

CP CV

or

C P = γ CV , therefore above expression

becomes

γ CV = CV + R (γ − 1) CV = R

R 8.314 = = 65.5 J mol −1 K −1 γ − 1 1.127 − 1 C P = CV + R = 65.5 + 8.314 = 73.8 J mol −1 K −1 CV =

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 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 3 CV = n R 2 2 CV 2(49.8) n= = ≅ 4 moles 3 R 5(8.314) (b) U = CV T = (49.8)(300) = 1.494 × 10 4 J 2 3 (c) C P = CV + n R = CV + CV Θ CV = n R 3 2 5 5 C P = CV = (49.8) = 83 J / K 3 3 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 J of heat are added at constant pressure. (c) Find the same quantities when 500 J of heat are added at constant volume. Solution (a) The initial internal energy is given by 3 U i = CV T = R T 2 3 U i = (8.314)(273) = 3405 J 2

CH 03 THE FIRST LAW OF THERMODYNAMICS 65 dQ (b) Now dQ = C P dT or dT = CP

 dQ  dQ 500  = dU = CV dT = CV  = = 300 J  C P  (C P / CV ) (5 / 3) Therefore the final internal energy and work are given by U f = U i + dU = 3405 + 300 = 3705 J dW = dQ − dU = 500 − 300 = 200 J (c) dQ = dU = 500 J and dW = p dV = p(0) = 0

CH 03 THE FIRST LAW OF THERMODYNAMICS 66 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 10 0 C at constant volume. Given that C v = 1.82 J g −1 0 C −1 . Solution According to first law of thermodynamics dQ = dU + dW m C v dT = dU + p dV

(10 g )(1.82 J g −1 0 C −1 )(10 0 C ) = dU + p(0) 182 J = dU Increase in internal energy = dE INTERNAL = dU = 182 J

CH 03 THE FIRST LAW OF THERMODYNAMICS 67 3-2(B) WORK DONE AT CONSTANT PRESSURE Problem 3-14 A gas expands at atmospheric pressure and its volume increases by 500 cm 3 . Calculate the work done by the gas. ( 1 Atmosphere = 1.013 × 10 5 N / m 2 ) B.U. B.Sc. 1998A Solution The work done by the gas at constant pressure is given by W = p dV W = (1.013 × 10 5 )(500 × 10 −6 ) = 50.65 J Problem 3-15 One kilogram water is converted to steam at standard atmospheric pressure. The volume changes from 1 × 10 −3 m 3 as a liquid to 1.671 m 3 as steam. For this process calculate the work done on the system when the pressure is 1.013 × 10 5 N / m 2 . K.U. B.Sc. 1998 Solution The work done by the vapourizing water is W = p dV = p (V f − Vi )

W = (1.013 × 10 5 )(1.671 − 1 × 10 −3 ) = 1.691 × 10 5 J Problem 3-16 A gas expands at atmospheric pressure and its volume increases by 334 cm 3 . Find the work done by the gas. The atmospheric pressure is 1.013 × 10 6 dynes / cm 2 . B.U. B.Sc. (Hons.) 1988A The work done by the gas, at constant pressure, is given by W = p dV = p (V f − Vi ) Now p = 1.013 × 10 6 dynes / cm 2 p=

(1.013 × 10 6 ) × 10 −5 N / m2 −2 2 (10 )

Θ 1 dyne = 10 −5 N

CH 03 THE FIRST LAW OF THERMODYNAMICS 68 p = 1.013 × 105 N / m 2 dV = 334 cm 3 = (334)((10 −2 ) 3 m 3 = 3.34 × 10 −4 m 3 Hence dW = (1.013 × 10 5 )(3.34 × 10 −4 ) = 33.834 J Problem 3-17 A gas is compressed at a constant pressure of 0.8 atm from 9 litres to 2 litres . In the process, 400 J 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 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 = dQ − dW dU = −400 − (−567.28) = 167.28 J

CH 03 THE FIRST LAW OF THERMODYNAMICS 69 3-2(C) WORK DONE AT CONSTANT TEMPERATURE Problem 3-18 A gram molecule of a gas at 77 0 C 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 Vf  W = n R T λn    Vi 

 2 V0   = 2017 J W = (1)(8.314)(77 + 273) λn  V0  Problem 3-19 Calculate the work done by an external agent in compressing 1.12 moles of oxygen from volume of 22.4 litres and 1.32 atm pressure to 15.3 litres at the same temperature. P.U. B.Sc. 2001 Solution The desired work on the gas by the external agent is given by Vf  Vf  W = n R T λn   = piVi λn    Vi   Vi   15.3  W = (1.32 × (1.013 × 10 5 )}(22.4 × 10 −3 ) λn    22.4  W = −1.142 × 10 3 J = −1.142 kJ

Problem 3-20 A sample of gas consisting of 0.11 moles is compressed from a volume of 4.0 m 3 to 1.0 m 3 while its pressure increases from 10 to 40 Pa ( N / m 2 ) . 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 piVi = p f V f = n R T = 40 N • m which indicates that the given process is isothermal. The work done at constant temperature is given by  p  W = n R T λn  i  = (40) λn (10 / 40) = −55.5 J p   f  Problem 3-21 A balloon contains 0.30 mol of helium. It rises, while maintaining at constant 300 K 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 Vf  W = n R T λn    Vi 

 5 V0   = 1.203 × 10 3 J W = (0.30)(8.314)(300) λn  V  0  Problem 3-22 One mole of nitrogen gas is compressed isothermally from 10 N / m 2 to 20 N / m 2 at 27 0 C . Calculate the work done. Solution The work done during this compression is given by  pf   W = −n R T λn  pi   20  W = −(1)(8.314)(27 + 273) λn  = −1729 J  10 

CH 03 THE FIRST LAW OF THERMODYNAMICS 71 3-2(D) WORK DONE IN THERMAL ISOLATION OR UNDER ADIABATIC CONDITIONS Problem 3-23 One mole of oxygen, initially kept at 17 0 C , 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 C v = 21.1 J mol −1 K −1 is for oxygen. Solution (a) For adiabatic process p f V fγ = piVi γ

p f p −f γ ( p f V f ) γ = pi pi−γ ( piVi ) γ p1f−γ (n R T f ) γ = pi1−γ (n R Ti ) γ p1f−γ T fγ = pi1−γ Ti γ (1−γ ) / γ

 p  T f = Ti  i  p   f  0 Now Ti = 17 C = (17 + 273) K = 290 K pi 1 1 − γ 1 − 1.40 2 = = 0.1 , γ = 1.40 and = =− p f 10 γ 1.40 7

Hence (b)

T f = (290)(0.1) −2 / 7 = 560 K dW = dE INTERNAL = m C v dT

dW = (1)(21.1)(560 − 290) = 5697 J 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 γ = 1.40 for air. Solution

CH 03 THE FIRST LAW OF THERMODYNAMICS 72 The initial volume V0 of air at S.T.P. is given by

p1V1 = n R T1 3

n R T1 V1 = p1 (1)(8.314)(0 + 273) V1 = V0 = = 2.24 × 10 − 2 m 5 1.013 ×10 The final pressure of air, under adiabatic conditions, is given by V p f = pi  i V  f

   

γ

1.40

 V0   = 2.673 × 10 5 Pa p f = (1.013 × 10 )  0 . 5 V 0   The desired work done is p f V f − piVi W= γ −1 (2.673 × 10 5 ){(0.5) × (2.24 × 10 −2 )} − (1.013 × 10 5 )(2.24 × 10 −2 ) W = 1 − 1.40 W = −1812 J 5

Problem 3-25 2 m 3 of a gas at 100 N / m 2 expands according to law P V 1.2 = C where C is a constant, until volume is doubled. Calculate the work done. Solution Now PiVi1.2 = C

C = (100)(2)1.2 = 229.7 Hence P V 1.2 = 229.7 229.7 p= V The work done by the gas is given by

CH 03 THE FIRST LAW OF THERMODYNAMICS 73 V2

W=

4

−1.2 ∫ p dV = 229.7 ∫ V dV

2

V1 4

 V   229.7  −0.2 W = 229.7  4 − 2 −0.2 = 129.43 J  =   − 0 .2  2  − 0 .2  Problem 3-26 Five litres of argon at 0 0 C 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 p f V fγ = piVi γ −0.2

[

 n R T f  γ  n R Ti   V =  V f  f  Vi    γ −1 T f V f = TiVi γ −1

]

 γ  Vi1 Θ p V = n R T 

γ −1

V  T f = Ti  i  V   f  T f = (273)(5 / 1) (5 / 3) −1 = 798 K

Problem 3-27 Calculate the final temperature of a sample of carbon dioxide of mass 16.0 g that is expanded reversibly and adiabatically from 500 cm 3 at 298.15 K to 2000 cm 3 . Solution For an adiabatic process V T f = Ti  i V  f

   

γ −1

 500  T f = (298.15)    2000  Problem 3-28

1.30 −1

= 196.71 K

CH 03 THE FIRST LAW OF THERMODYNAMICS 74 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 p f V fγ = piVi γ

 n R T f  γ  n R Ti   V =  V f  f  Vi    T f V fγ −1 = TiVi γ −1 Vf   Vi

  

γ −1

=

 γ  Vi1 Θ p V = n R T 

Ti Tf 1 /(γ −1)

T  = i  Vi  T f  V f  T 1 /(1.40−1) = = 0.177  Vi  2 T  V f = 0.177 Vi Vf

Problem 3-29 A 1.00 mol sample of an ideal diatomic gas originally at 1.00 atm and 20 0 C , expands adiabatically to twice its volume. What are final pressure and temperature for the gas? Assume no molecular vibrations. Solution For an adiabatic process p f V fγ = piVi γ V p f = pi  i V  f

   

γ

CH 03 THE FIRST LAW OF THERMODYNAMICS 75 1.40

 V  p f = (1) 0   2 V0  p f = 0.379 atm

Θ γ = 1.40 for a diatomic gas.

Further T f V fγ −1 = TiVi γ −1 V T f = Ti  i V  f T f = 222 K

   

γ −1

 V  = (20 + 273) 0   2 V0 

1.40 −1

Problem 3-30 An ideal gas initially at 8.00 atm and 300 K 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 V p f = pi  i V  f

   

γ

γ −1

V  T f = Ti  i  and V   f  5 (a) For monoatomic gas γ = therefore 3 5/3

 V p f = (8.00)  i  2 Vi

  

 V  T f = (300)  i   2 Vi 

( 5 / 3 ) −1

= 2.52 atm = 189 K

CH 03 THE FIRST LAW OF THERMODYNAMICS 76 7 (b) For diatomic gas γ = = 1.40 therefore 5

 V p f = (8.00)  i  2 Vi

  

 V T f = (300)  i  2 Vi

  

1.40

= 3.03 atm

1.40 −1

= 227 K

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 V T f = Ti  i V  f

   

γ −1

1.40

 V  T f = (0 + 273)  0  = 176 K  3 V0  The final pressure of the gas is given by V p f = pi  i V  f

   

γ

1.40

 V  p f = (1.013 × 10 )  0  = 2.176 × 10 4 Pa  3 V0  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 (b) Adiabatic. K.U. B.Sc. 2000 5

CH 03 THE FIRST LAW OF THERMODYNAMICS 77 Solution (a) For an isothermal process p f V f = p iVi Vf

pi pi = =3 Vi p f (1 / 3) p i (b) For an adiabatic process p f V fγ = piVi γ Vf   Vi

=

γ

 p  = i pf   p  = i  Vi  p f 

Vf

1/ γ

1 /( 5 / 3 )

 pi  =   (1 / 3) pi 

= (3) 0.6 = 1.933

Example 3-33 A volume of argon gas at 27 0 C expands adiabatically until its volume is increased four times. Find the resulting fall in temperature. Given that γ = 1.67 is for argon gas. Solution V Now T f = Ti  i V  f

   

γ −1

1.67 −1

 V  T f = (27 + 273)  0  = 118.5 K or − 154.50 C  4 V0  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? (b) What is the final temperature of the gas if it is diatomic?

CH 03 THE FIRST LAW OF THERMODYNAMICS 78 Solution

Now

V T f = Ti  i V  f

   

γ −1

(a) For monoatomic gas γ =

5 therefore 3 ( 5 / 3 ) −1

 V0   T f = (300)  = 476 K 0 . 5 V 0   7 (b) For diatomic gas γ = = 1.40 therefore 5  V0 T f = (300)   0.5 V0

  

1.40 −1

= 396 K

Example 3-35 Calculate the rise in temperature when a gas at 27 0 C 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  p  T f = Ti  i  p   f 

(1−γ ) / γ

(1−1.5 ) / 1.5

 p  T f = (27 + 273) i  = 600 K  8 pi  The rise in temperature of the gas will be ∆T = T f − Ti = 600 − 300 = 300 K Example 3-36 A given mass of gas at 0 0 C is suddenly compressed to a pressure twenty times the initial pressure. What will be the final temperature of the gas if γ is 1.42 .

CH 03 THE FIRST LAW OF THERMODYNAMICS 79 Solution The final temperature of the gas is given by the expression  p  T f = Ti  i  p   f 

(1−γ ) / γ

 pi T f = (0 + 273)  20 pi

  

(1−1.42 ) / 1.42

= 662.2 K

Example 3-37 One mol of an ideal monoatomic gas at 300 K and 3.0 atm expands adiabatically to a final pressure of 1.0 atm . How much work does the gas do in the expansion? Solution For adiabatic process  p  T f = Ti  i  p   f 

(1−γ ) / γ

(1−1.67 ) / 1.67

 3 .0  T f = (300) = 193 K   1 .0  The work done in adiabatic process is given by R 8.314(300 − 193) W= (Ti − T f ) = = 1328 J γ −1 1.67 − 1 Example 3-38 An ideal monoatomic gas, consisting of 2.6 mol of volume 0.084 m 3 , expands adiabatically. The initial and final temperatures are 25 0 C and − 68 0 C respectively. What is the final volume of the gas? Solution For adiabatic process T f V fγ −1 = TiVi γ −1

CH 03 THE FIRST LAW OF THERMODYNAMICS 80 T V fγ −1 =  i T  f T Vf =  i T  f

 γ −1  Vi      

1 /(γ −1)

Now Vi = 0.084 m 3 Ti = 25 0 C = (25 + 273) K = 298 K

T f = −68 0 C = (−68 + 273) K = 205 K

γ = 1.67 for monoatomic gas. Hence  298  V2 =    205 

1 /(1.67 −1)

(0.084) = 0.147 m 3

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. ( 1 calorie = 4.18 J ) B.U. B.Sc. 2007A (2) The atmospheric pressure is 1.013 × 10 6 dynes / cm 2 . 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 by 400 cm 3 . 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 27 0 C and γ = 1.5 . F.P.S.C. 1978

Answers (1) 418 J (4) 600 K

(2) 67.67 J

(3) 40.52 J