2924990-bansal-pattern-thermodynamics-2-solution-by-SKsinha-See-Chemistry-Animations-at-sinhalabcom[1].pdf...

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SINHA ’ s I. I.T. CHEMISTRY - 1 -

1-D-10 Talwandi & B-7. JAWAHAR NAGAR, KOTA.(RAJ) Ph-0744-2422383 .Mo-93149-05055

THERMODYNAMICSProblems

1. 1.25 m 3 of air at 180 0C at 8 bar is undergoing a constant pressure until the volume is doubled. Determine thechange in the entropy and enthalpy of air. , Given C P = 1.005 kJ/kg

0K, C v = 0.718 kJ/kg0K and R = 0.287 kJ/kg

0K

1 Solution: Assuming air behaves like a perfect gas we have, C P = 1.005 kJ/kg 0K,Cv = 0.718 kJ/kg

0K and R = 0.287 kJ/kg 0KGiven: V 1 = 1.25 m

3, T 1 = 180 + 273 = 4530 K, P 1 = P 2 = 8 x 10

5N/m 2 and V 2 = 2V 1 For a constant pressure process, change in entropy is

!! " #

$$% & '(

1

212 ln

T

T mC S S P

We have P 1V1 = mRT 1

kg x x x

m 69.7453287

25.1108 5 '')

2

22

1

11

T

V P

T

V P Also '

K x x

T 02 90625.1

45325.12 '')

Therefore change in entropy =453

906ln005.169.7 x

= 5.596 kJ/ 0KChange in enthalpy = Q 1-2 = mC P (T 2 – T1) = 3500.99 kJ

2. 1kg of air initially at 27 0C is heated reversibly at constant pressure until the volume is doubled, and then is heatedat constant volume until the pressure is doubled. For the total path find i) Work transfer, ii) Heat transfer, iii)Change in entropy

2. Solution: Given: m = 1 kg, T 1 = 3000 K V 2 = 2V 1, P 3 = 2P 2 = 2P 1

Process 1-2: Constant pressure processi) Work done, W 1-2 = P (V 2 – V 1)

= PV 2 – PV 1 = mR (T 2 – T1)

2

22

1

11

T

V P

T

V P Also '

But P 1 = P 2 122

2

1

1 2V V But T

V

T

V '')

Therefore T 2 = 2T 1 = 6000K

Therefore work done W 1-2 = 1 x 0.287 x (600 – 300) = 86.1 kJ

V

P

3

21





kg x

xm 128.0

533287.0

028.0700 '')

3027.0

084.0

11

22

1

2 '''V P

V P

T

T Now

Therefore T 2 = 3 x 533 = 1559 K

27)3(533

1599 35.05.1

1

3

2

3

2 ''! " #$

% & '!! "

#$$% & '

(nn

T

T

P

P Again

Heat transfer in process 1-2, Q 1-2 = mc p (T 2 – T 1)= 0.128 x 1.005 (1599 – 533)= 137.13kJ

Heat transfer in process 2 -3,

pdvU Q *+,'( 32

- . - .1

3223 (

(+('n

T T mRT T mc v

- .231

T T nn

mc v (((' /

On substituting Q 2-3 = - 19.59 kJ

For process 3-1Q3-1 = dU + W 3-1

But dU = 0, i.e., * '''' ((1

3 1

31

3

111313 lnln P

PmRT

v

vmRT pdvW Q

On substituting, Q 3-1 = - 64.53 kJ

Heat received in the cycle 137.13 kJ

Heat rejected in the cycle Q 2 = 19.59 + 64.53 = 84.12 kJ

The efficiency of the cycle

%3939.013.137

12.8411

1

2 or Q

Qcycle

'('('0

4. 1 kg of air at a pressure of 7 bar and a temperature of 90 0C undergoes a reversible polytropic process which may berepresented by PV 1.1 = C, final pressure is 1.4 bar. Evaluate i) The final specific volume, temperature and increasein entropy, ii) Work done and heat transfer.

4 Solution: Given, m = 1 kg, P 1 = 7 bar, T 1 = 3630K, PV 1.1 = C, P 2 = 1.4 bar

Air is perfect gas i.e., P 1V1 = mRT 1

kgm x

v / 14883.0107

)363)(287(1 351

'')

Also we have, P 1V11.1 = P 2V21.1

Therefore V 2 = 0.6429 m3

Also P 2V2 = mRT 2

Therefore T 2 = 313.610K

Change in entropy for a polytropic process is,





!! " #

$$% &

!! " #

$$% &

(('(

1

212 ln.1 V

V R

nS S

/

/

Substituting the values, noting / = 1.4 and R = 0.287 kJ/kg 0K, we getS2 – S1 = 0.31495kJ/kg 0K

Work done,1

221121 (

('(n

V PV PW

Substituting we get, W 1-2 = 141.75 kJ/kg

Heat transfer Q 1-2 = W 1-2 + (U 2 – U1)= W 1-2 + mC v (T 2 – T1)

= 141.75 – 35.462= 106.29 kJ/kg

5. Show that for a reversible adiabatic process the equation is Pv / = Constant.5 Solution: The general property relations for an ideal gas may be written as

Tds = du + pdv= c vdT + pdv

And, also Tds = dh – vdp= c pdT – vdp

For a reversible adiabatic change, ds = 0Therefore c v dT = - pdvAnd c p dT = vdpBy division,

pdvvdp

c

c

v

p ('' /

0'+v

dv pdp

Or /

Or d (ln p) + /d (ln v) = d (ln c) where c is the constantTherefore ln p + / ln v = ln ci.e., pv / = constant

6. A mass of 0.25 kg of an ideal gas has a pressure of 300 kPa, a temperature of 80 0C, and a volume of 0.07 m 3. Thegas undergoes an irreversible adiabatic process to a final pressure of 300 kPa and final volume of 0.1 m 3 duringwhich the work done on the gas is 25 kJ. Evaluate the c p and c v of the gas and increase in entropy of the gas.

6 Solution: We have, P 1 V 1 = mRT 1

kgK kJ x

x R / 238.0

35325.0

07.0300 '')

Final temperature K x x

mR

V PT 505

238.025.01.030022

2 ''' We have from first law of TD, Q – W = U 2 – U1

But Q = 0 Therefore W = U 1 – U2 W = mc v (T 1 – T 2)

On substituting, -25 = 0.25 x c v x (353 – 505)Therefore c v = 0.658 kJ/kg-K

Now R = c p – cv Therefore c p = 0.896 kJ/kg-K

Entropy change1

2

1

212 lnln

V

V mc

P

PmcS S pv +'(





07.0

10.0ln896.025.0ln

1

2 xV

V mc p ''

= 0.08 kJ/kg-K

7. An ideal gas cycle consisting of three processes uses Argon (M = 40) as a working substance. Process 1-2 is areversible adiabatic expansion from 0.014m 3, 700 kPa, 280 0C to 0.056m 3. Process 2-3 is a reversible isothermalprocess and process 3-1 is a constant pressure process. Sketch the cycle on P-v and T-s diagrams and determine (i)work transfer in each of the three processes (ii) heat transfer in each of the three processes and (iii) net work outputfrom the cycle. Assume for Argon / = 1.67.

7 Solution: M = 40, V 1 = 0.014m3, P 1 = 700 kPa, T 1 = 553 K

V2 = 0.056m3, / = 1.67

Figure P-v and T-s diagrams

Process 1-2 Reversible adiabatic process

K kgkJ M R

R (''' / 208.0403143.8

- . K kgkJ

Rcv ('(

'(

' / 311.0167.1

208.01/

cp = 0.208 + 0.311 = 0.519 kJ/kg-KSince the process is reversible adiabatic, the working substance is treated as an ideal gas

kPaPor

xV

V PPor V PV Pei

13.69

056.0

014.0700.,.

2

67.1

2

1122211

'

! " #

$% &

'!! "

#$$%

& ''

/

/ /

K xV

V T T Also 45.218

056.0

014.0553

67.01

2

112

'! " #$

% & '!! "

#$$%

& '

(/

- .- .

- . J

x x x xV PV PW

167.1

056.01013.69014.010700

1

331211

21 (('

(('(

/

= 8.85 x 10 3J = 8.85 kJQ1-2 = 0 as the process is reversible adiabatic.

Process 2-3 Reversible isothermal processT2 = T 3 = 218.45 K

!! " #

$$% & '!! "

#$$% & '!! "

#$$% & '(

1

22

3

22

2

3232 lnlnln

P

PmRT

P

PmRT

V

V mRT W

kg x x

x x RT

V Pm But 2157.0

45.21810208.0014.010700

3

3

1

11 '''





! " #$

% & ') (

700

13.69ln45.21810208.02157.0 332 x x x xW

= - 22.69 x 10 3J = - 22.69 kJBy first law of TD Q 2-3 = (U 3 – U 2) + W 2-3

But U 3 – U 2 = mc v (T 3 – T2) = 0Therefore Q 2-3 = W 2-3 = - 22.69 kJ

Process 3-1 Constant Pressure ProcessW 3-1 = (P 1V1 – P3V3) = mR (T 1 – T 3)

= 0.2157 x 0.208 x 10 3 (553 – 218.45) = 15 kJ

Therefore Net work output from the cycle = W c = W 1-2 + W 2-3 + W 3-1 = 8.85 – 22.69 + 5.93= - 7.91 kJ

For the cyclic process we haveW QQQor W Q 1'++1'1 ((( 133221

0 - 22.69 + Q 3-1 = - 7.91

Therefore Q 3-1 = 14.78 kJ

8. A mass of an ideal gas exists initially at 200 kPa, 300K and 0.5 m 3 / kg. The value of / is 1.4. (i)Determine thespecific heats of the gas. (ii)What is the change in entropy when the gas is expanded to a pressure of 100 kPaaccording to the law pv 1.3 = constant (iii) What will be the change in entropy if the process is according to the lawpv1.5 = constant (iv) What inference you can draw from this example.

8 Solution: P 1 = 200 kPa, T 1 = 300 K, v 1 = 0.5 m3 /kg, / =1.4, P 2 = 100 kPa

(i) For an ideal gas P 1v1 = RT 1 or- .

K kg J T

vP R ('' / 33.333

1

11

K kg J cc

K kg J R

c

v p

v

(''

('(

'

/ 66.1166

/ 33.8331

/

/

(ii) For the given process nn vPvP 2211 '

kgmvP

Pvor

n

/ 852.0 31

1

2

12 '!! "

#$$% & '

- .K

R

vPT Also 6.255222 ''

Change in entropy per unit mass is given by

!! " #$$% & +!! " #$$% & '( 12

1

212 lnln

vv R

T T cS S v

= 44.18J/K

(iii) For the process for which the index n = 1.5 we have

kgm xv / 764.05.0100

200 35.11

2 '! " #$

% & '





K x x

T 2.23833.333

794.010100 32 ''

! " #$

% & +!

" #$

% & '()

5.0

794.0ln33.333

300

2.238ln33.83312 x xS S

= - 38.07 J/K

(iv) It can be seen from the results of (ii) and (iii) above that entropy increases when n < / and it decreases when n> / .

9. A closed rigid cylinder is divided by a diaphragm into two equal compartments, each of volume 0.1 m 3. Eachcompartment contains air at a temperature of 20 0C. The pressure in one compartment is 2.5 MPa and in the othercompartment it is 1 MPa. The diaphragm is ruptured so that the air in the compartments mix to bring the pressure toa uniform value through the insulated cylinder. Find the net change in entropy due to mixing process.

9 Solution: Let suffix 1 denote the condition of air in the left half of the cylinder, suffix 2 the conditions in the righthalf of the cylinder and suffix 3 the condition after mixingIt if given that,

V1 = V 2 = 0.1 m3, T 1 = T 2 = 20 + 273 = 293K, P 1 = 2.5 MPA, P 2 = 1.0 MPa

After mixing the temperature is given by

2211

2221113

p p

p p

cmcm

T cmT cmT +

+'

Since T 1 = T 2 and c p1 = c p2, it follows that T 3 = T 1 = T 2 = 293 K(, S)universe = ( , S)1 + ( , S)2 + ( , S) surroundings

(, S) surroundings = 0 as the cylinder is insulated and no heat transfer can take place during the mixing process.Therefore ( , S)universe = ( , S)1 + ( , S)2

234

567 ++2

34

567 +

2

3

2

32

1

3

1

31 lnlnlnln

v

v R

T

T cm

v

v R

T

T cm vv

kg x

x x RT

vPm 973.2

2932871.0105.2 6

1

111 '''

kg x

x xmSimilarly 189.1

2932871.0100.1 6

2 '' Therefore on substituting, ( , S)universe = 0.828kJ/K

10. Air is compressed from 1 bar, 27 0C to 4.5 bar, 177 0C. Determine the change in entropy per kg of air (i) takingvariation of specific heats into account and (ii) assuming specific heat at constant pressure to be constant and isequal to 1.01 kJ/kg-K over this range of temperature.

10 Solution: T 1 = 300 K, T 2 = 450 K, P 1 = 1 bar, P 2 = 4.5 bar

(i) We have change in entropy per kg of air, - . !! " #

$$% & (('(

1

201

0212 ln p

p Rssss

From table C-21 at T 1 = 300 K, K kgkJ s (' / 5153.201 and

at T 2 = 450 K, K kgkJ s (' / 9245.202 Therefore on substituting, s 2 – s1 = - 0.02508 kJ/kg-K

(ii) The change in entropy per kg of air

1

2

1

212 lnln

P

P R

T

T css p ('(





K kgkJ (('(' / 0252.01

5.4ln289.0

300

450ln01.1

11 a) A gaseous mixture contains 21% by volume of N 2, 50% by volume of H 2 and 29% by volume of CO 2. Calculatethe gas constant of the mixture and the ratio of specific heats. If the mixture pressure is at 1 bar and the mixturetemperature is 10 0C, calculate the partial pressures and mass fractions of the constituents.

b) A cylinder contains 0.085m 3 of this mixture at 1 bar and 10 0C. The gas undergoes a polytropic process accordingto the law Pv 1.2 = constant to a final volume which is one fifth of the initial volume. Determine i) the magnitude anddirection of work transfer, ii) magnitude and direction of heat transfer and iii) the change in entropy.

11 Solution: From tables C-6, M N2 = 28.02, M H2 = 2.016 and M CO2 = 44From table C-11, C PN2 = 1.039 kJ/kg-K, C PH2 = 14.15 kJ/kg-K and C PCO2 = 0.818 kJ/kg-Ka) Given V N2 / V = V fN2 = 0.21, V H2 / V = V fH2 = 0.5, V CO2 / V = V fCO2 = 0.29

P = 1 bar, T = 283 K

K kgkJ M

R R

K kgkJ M

R R

K kgkJ M

R RhaveWe

COCO

H

H

N N

('''

('''

('''

/ 189.0443143.8

/ 124.4016.2

3143.8

/ 297.0013.28

3143.8

22

2

2

22

Therefore C vN2 = 1.039 – 0.297 = 0.742 kJ/kg-KCvH2 = 14.15 – 4.124 = 10.026 kJ/kg-KCvCO2 = 0.818 – 0.189 = 0.629 kJ/kg-K

We know that volume fraction = mole fractioni.e., y N2 = V fN2 = 0.21

yH2 = V fH2 = 0.5yCO2 = V fCO2 = 0.29

Molecular weight of the mixture M m = y N2 M N2 + y H2 M H2 + y CO2 M CO2 = 0.21 x 28.013 + 0.5 x 2.016 + 0.29 x 44

= 19.654 kg/kg-moleGas constant for the mixture = R m = 8.3143 / 19.654

= 0.423 kJ/kg-K

Mass fraction of N 2 m

N N

mm

N N

m

N fN M

M y

M n

M n

m

mm 22

2222 '''

2993.0654.19

013.2821.0 '' x

Similarly 0513.0654.19

016.25.02 '' xm fH

6494.0

654.19

4429.02 '' xm fCO

For the mixture C vm = M fN2 C vN2 + M fH2 C vH2 + M fCO2 C vCO2 = 0.2993 x 0.742 + 0.0513 x 10.026 + 0.6494 x 0.629= 1.145 kJ/kg-K

CPm = C v + R m = 1.145 + 0.423

= 1.568 kJ/kg-KTherefore the ratio of specific heat for the mixture is /m = C Pm / C vm

= 1.568 / 1.145= 1.369





b) Mass of the mixture kg x x x

RT

V pm 071.0

283423085.0101 5

1

11 ''!! " #

$$% & ''

For the given process, 2.1222.1

11 V pV p '

- . bar x pV V

p 9.615 2.1

1

2.1

2

1

2 ''!! "

#$$%

& ')

- .1

22112121 (

('' (( nV pV p

W W d

- . J

x x x x x

16150

12.1

085.05

1109.6085.0101 55

('(

234

567

! " #$

% & (

'

Negative sign indicates that work is done by the surroundings on the gas mixture.Temperature of the mixture at the end of the process is,

K x

x x x

mR

V pT

6.390423071.0

085.051

109.6 522

2

'

! " #

$% &

''

Or K T ei p

p

T

T nn

5.39019.6

283.,.2.1

2.0

2

1

1

2

1

2 '! " #$

% & '!! "

#$$% & '

(

Change in internal energy, U 2 – U1 = mC v (T 2 – T 1)= 0.071 x 1.145 x (390.6 – 283)= 8.747 kJ

By first law of TD, Q 1-2 = W 1-2 + (U 2 – U 1)= - 16.15 + 8.747= - 7.403 kJ

Negative sign indicates that heat transfer takes place from the mixture to the surroundings

Change in entropy of the mixture 23

456

7!! "

#$$%

& +!! "

#$$%

& '('

1

2

1

212 lnln

V

V R

T

T C mS S v

K kJ

x x

/ 02214.0

5

1ln423.0

283

6.390ln145.1071.0

('

234

567

! " #$

% & +!

" #$

% & '

12. Calculate the constant volume and constant pressure specific heats of a gas mixture consisting of 1 kg ofoxygen and 2 kg of nitrogen at a pressure of 1.5 bar and temperature 20 0C. Also determine the change in internalenergy, enthalpy and entropy of the mixture when it is heated under constant volume to a temperature of 100 0C.

12 Solution: From tables C-6, M O2 = 32, M N2 = 28From table C-11, C PN2 = 1.038 kJ/kg-K, C PO2 = 0.917 kJ/kg-K

) CvN2 = C pN2 – RN2 = 1.038 – 8.3143 / 28 = 0.741 kJ/kg-KCvO2 = C pO2 – RO2 = 0.917 – 8.3143 / 32 = 0.653 kJ/kg-K

- . - .1'i

iv fimv cmc


http:///reader/full/2924990-bansal-pattern-thermodynamics-2-solution-by-sksinha-see-chemistry-animations-at-si… 10/13



K kgkJ x x ('+' / 712.0741.03

2653.0

3

1

- . - .1'i

i p fim p cmc

K kgkJ x x ('+' / 998.0038.13

2917.0

3

1

Change in internal energy,

- . - .12 T T cmU mvm (', = 3 x 0.712 (100 – 20)= 170.88 kJ

Change in enthalpy,

- .12 T T cm H m pm (', = 3 x 0.998 (100 – 20)= 239.52 kJ

Change in entropy of the mixture,

- . - . 23

456

7

!! "

#

$$%

& +

!! "

#

$$%

& ',

1

2

1

2 lnlnV

V R

T

T cmS

mmvmm

Since the volume is constant, the change in entropy,

- . - . 23

456

7!! "

#$$%

& ',

1

2lnT

T cmS mvmm

293

373ln712.03 x'

= 0.5156 kJ/K

13. A gaseous mixture consisting of 1 kg of helium and 2.5 kg of nitrogen is compressed isentropically in a closedsystem from a pressure of 1 bar, 27 0C to a pressure of 7 bar. Assuming specific heats of helium and nitrogen tobe constant, determine the specific heats of the mixture, the change in entropy of individual gases, and thechange in internal energy of the mixture. Also find the work required for the compression process. AssumeCvHe = 3.14 kJ/kg-K, C vN2 = 0.741 kJ/kg-K, C pHe = 5.233 kJ/kg-K and C pN2 = 1.038 kJ/kg-K

13 Solution: - . - .1'i

iv fimv cmc

K kgkJ x x ('+' / 426.1741.05.3

5.2140.3

5.3

1

- . - .1'i

i p fim p cmc

K kgkJ x x ('+' / 236.2038.15.3

5.2233.5

5.3

1

- .- .

568.1426.1236.2 ''')

mv

m p

m c

c/

m

m

p

pT T

/

/ 1

1

212

(

!! " #

$$% & ')

K 6071

7300

568.1

568.0

'! " #$

% & '





Hence the change in entropy of helium

- . - .223

4

556

7(',

1,

2,

1

2 lnln He

He He He p He He p

p R

T

T cmS

1

7,

1

2

1,

2,

1,

2,

2

2 ''' p

p

p

p

p

pthat knowWe

N

N

He

He

- . 234

567 (',)

1

7ln093.2

300

607ln233.51 HeS

= - 0.38 kJ/K

Similarly - . 234

567 (',

1

7ln297.0

300

607ln038.15.2

2 N S

= + 0.38 kJ/KInternal energy of the mixture

- . - .12 T T cmU mvmm (', = 3.5 x 1.426 (607 – 300)= 1532.24 kJ

For a closed system we have, ä q – ä w = du.Since the process adiabatic, ä q = 0Therefore ä w = -dU.Therefore the work required for the compression process,

W 1-2 = - (dU) m = - 1532.24 kJ

14. Two thermally insulated vessels, each of 0.85 m 3 in volume are isolated from each other by a partition wall. One ofthe vessels contains nitrogen and the other oxygen, each at 5 bar and 100 0C. As soon as the partition between thevessels is removed, the two gases mix adiabatically. Determine the increase in entropy of the gas mixture.

14. Solution: Since the pressure and temperature of each gas are equal, the pressure and temperature of the mixturewill not change, i.e., the mixture will be at 5 bar and 373 K. But the volume of the mixture will be 2x0.85= 1.7m 3 andthe pressure of N 2 and O 2 is

21

7.185.0

22 '''m

N

m

N

v

v

p

p

Therefore p N2=0.5x5=2.5 bar and p O2=2.5 barUsing perfect gas equation, mass of the N 2,

kg

x x

x xm N 84.3

37328

103143.8

7.1105.23

5

2''

and kg

x x

x xmO 39.4

37332

103143.8

7.1105.23

5

2''

Therefore the increase in entropy of the mixture is,- .222222

lnln OOO N N N m y Rm y RmS +(',

234

567 +('

5

5.2ln

32

3143.839.4

5

5.2ln

28

3143.884.3 x x

=1.579kJ/K

15. Two kg-mole of CO 2 at a pressure of 1.8 bar, 80 0C is mixed in a thermally insulated vessel with 3 kg-mole of N 2 at2.2 bar, 60 0C. When the mixture is at equilibrium, determine the final temperature and pressure and the change inentropy of the mixture. Assume C vCO2 = 0.653 kJ/kg-K, C vN2 = 0.741 kJ/kg-K





15 Solution: we have,- . - .

- .mvm N N v N COCOvCO

m cn

T cnT cnT 222222

+'

Now the molal specific heats at constant volume of the constituent gases are

- . - .222 COvCOCOv

c x M c ' = 44x0.653=28.73 kJ/kg-mole K

- . - .222 N v N N v

c x M c And ' =28x0.741=20.75 kJ/kg-mole K

75.20373.282

33375.20335373.282

x x x x x x

T m ++')

=342.6 KThe final pressure of the mixture is

m

mm

m V

T Rn p '

The volume of the individual gases is,

361.328.11003533143.82

2

22

2m

x x x

p

T RnV

CO

COCOCO '''

375.372.21003333143.83

2

22

2m

x x x

p

T RnV

N

N N N '''

336.7075.3761.3222

mV V V N COm '+'+') Therefore the final pressure of the gas mixture is,

bar x

x x pm 02.236.701006.3423143.85 ''

The partial pressure of the individual gases,

bar x xpn

n p m

m

COCO 808.002.25

222

'''

bar x xpn

n p m

m

N N 212.102.25

322

'''

Therefore the entropy change of the mixture,

- . - . - .22 N COm

S S S ,+,',

- .2234

5567 ('

1,

2,

2

2

2

22lnln

CO

CO

CO

mCOvCO p

p RT T cn

- .223

4

556

7(+

1,

2,

2

2

2

22lnln

N

N

N

m N v N p

p R

T

T cn

- . - . 73.283143.822

+'+'COvCO p

c Rc But

=37.04 kJ/kg mole K





- . - . 75.203143.822

+'+' N v N p

c Rcand

=29.064 kJ/kg mole K

- . +234

567 (',)

8.1

212.1ln3143.8

353

6.342ln04.372mS

234567 ( 2.2212.1ln3143.83336.342ln064.295

=2[-1.108+6.659]+3[0.826+4.957]=28.45 kJ/ K

2924990-bansal-pattern-thermodynamics-2-solution-by-SKsinha-See-Chemistry-Animations-at-sinhalabcom[1].pdf...

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