Tutorial 2

ByMiss Anis Atikah Ahmad

Question 1

A heat engine uses reservoirs at 800°C and 0°C.a) Calculate maximum possible efficiencyb) If qH is 1000 J, calculate the maximum -w and the

minimum value of -qC

Question 1

a) Calculate maximum possible efficiency

TH= 800°C = 1073.15K, TC =0°C =273.15K

rev Work output per cycle

Energy input per cycle H

C

H

C

T

T

q

q 11

745.015.1073

15.2731

K

K

Question 1

b) If qH is 1000 J, calculate the maximum -w and the minimum value of -qC

rev Work output per cycle

Energy input per cycle

J

w

1000745.0

JJw 7451000745.0max

H

C

q

q1

H

C

q

q1745.0

J

qC1000

1745.0

JqC 1000745.01000 JqC 255

Question 2

Calculate ΔS for each of the following changes in state of 2.50 mol of a perfect monoatomic gas with CV,m =1.5R for all temperatures:a) (1.50 atm, 400K) (3.00 atm, 600K)b) (2.50 atm, 20.0L) (2.00 atm, 30.0L)c) (28.5L, 400K) (42.0L atm, 400K)

Question 2

• For perfect gas;

revrev dwdUdq

dVTnRTdTCPdVdTCdq VVrev

revrev dwdqdU TqddS rev

VnRdVTdTCdS V

2

1

2

1

V

V

T

T

V VnRdVTdTCS

Question 2

• Since CV,m is constant (same for all temperatures) ;

• Given, , thus:

1212 lnln VVnRTTCS V

RC mV 5.1,

1212, lnln VVnRTTCn mV

1212 lnln5.1 VVnRTTnRS

1212 lnln5.1 VVTTnR

Question 2

a) P1 = 1.5 atm, P2= 3 atm, T1 = 400 K, T2= 600K

1212 lnln5.1 VVTTnRS

KatmatmKKmolJmol 40035.1600ln400600ln5.1314.85.2

1212ln400600ln5.1314.85.2 TPPTKmolJmol

KJ66.6

Question 2

b) P1 = 2.5 atm, P2= 2 atm, V1 = 20L, V2= 30L

1212 lnln5.1 VVTTnRS

LLLatmLatmKmolJmol 2030ln205.2302ln5.1314.85.2

LLVPVPKmolJmol 2030lnln5.1314.85.2 1122

KJ1.14

Question 2

c) V1 = 28.5L, V2= 42L, T1= T2 = 400K

1212 lnln5.1 VVTTnRS

LLKKKmolJmol 5.2842ln400400ln5.1314.85.2

KJ06.8

Question 3After 200 g of gold [cP = 0.0313 cal/(g °C)] at 120°C is dropped into 25.0 g of water at 10°C, the system is allowed to reach equilibrium in an adiabatic container. Calculate:

a) The final temperatureb) ΔSAu

c) ΔSH2O

d) ΔSAu + ΔSH2O

Question 3

• At constant pressure;

• At equilibrium;

dTmcdTCdqdq PPPrev

2

1

T

T

rev

T

dqS

122,2121,1 TTcmTTcm PP

101200313.0200 22,22 TcmTCgcalg P

101200313.0200 22,22 TcmTCgcalg P

(a)

Question 3

• Solving for T2;

CT 322

1000.1251200313.0200 22 TCgcalgTCgcalg

102512026.6 22 TCcalTCcal

2502526.62.751 22 TT

2.100126.31 2 T

Question 3

2

1

T

T

P

T

dTmcS

2

1

T

T

rev

T

dqS

(b)dTmcdTCdqdq PPPrev

12ln TTmcS P

15.39315.305ln/0313.0200 KgcalgS

Kcal /59.1

Question 3

2

1

T

T

P

T

dTmcS

2

1

T

T

rev

T

dqS

(c)dTmcdTCdqdq PPPrev

12ln TTmcS P

15.28315.305ln/00.125 KgcalgS

Kcal /87.1

Question 3(d)

waterAutotal SSS

KcalKcal /28.0/87.159.1

Question 4

A sample consisting of 2.00 mol of diatomic perfect gas molecules at 250 K is compressed reversibly and adiabatically until its temperature reaches 300 K. Given that CV,m = 27.5 JK-1mol-1, calculate:a) qb) wc) ΔUd) ΔHe) ΔS.

n = 2.00 mol CV,m = 27.5 JK-1mol-1

T1= 250 K T2 = 300 K

Process: reversible adiabatic of a perfect gas

a) qrev =0

b) w = ?Recall first law:ΔU = q + wFor perfect gas,

Thus,

Question 4

wdTCV 0 JKmolKJmoldTCV 27502503005.270.2 1

dTCU V

Jw 2750

Question 4c) ΔU ΔU = q + w = 0 + w = w = 2750 J

d) ΔHFor a perfect gas;

is not given. However, we know that,Thus;

dTCH P

PC nRCC VP

dTCnRH V

dTnCnR mV ,

KKJmolKJmol 2503005.27314.80.2 111

J4.3581

Question 4e) ΔS

2

1 T

dqS rev

Since qrev =0 (reversible adiabatic),

0S

Question 5 A system consisting of 1.5 mol CO2 (g), initially at 15°C and

9 atm and confined to a cylinder of cross-section 100.0 cm2. It is allowed to expand adiabatically against an external pressure of 1.5 atm until the piston has moved outwards through 15 cm. Assume that carbon dioxide may be considered a perfect gas with CV,m = 28.8 JK-1mol-1, calculate:a) qb) wc) ΔUd) ΔTe) ΔS

A system consisting of 1.5 mol CO2 (g), initially at 15°C and 9 atm and confined to a cylinder of cross-section 100.0 cm2. It is allowed to expand adiabatically against an external pressure of 1.5 atm until the piston has moved outwards through 15 cm. Assume that carbon dioxide may be considered a perfect gas with CV,m = 28.8 JK-1mol-1, calculate:a) qb) wc) ΔUd) ΔTe) ΔS

Question 5

15 cm

V1 V2

T1 = 15°C

P1 = 9 atm

Pext = 1.5 atm

A = 100 cm2

• n = 1.5 mol CO2

• CV,m = 28.8 JK-1mol-1

• A perfect gas• Adiabatic

a) q = 0 (adiabatic)

b) w = ?

Question 5

15 cm

V1 V2

T1 = 15°C

P1 = 9 atm

Pext = 1.5 atm

A = 100 cm2

VPW ext

cmcmatm 151005.1 2

Jcm

m

atm

Pacmatm 25.227

10

1

1

1001.12250

36

353

• n = 1.5 mol CO2

• CV,m = 28.8 JK-1mol-1

• A perfect gas• Adiabatic

c) ΔU = ?

d) ΔT = ? For perfect gas, Thus,

Question 5

15 cm

V1 V2

T1 = 15°C

P1 = 9 atm

Pext = 1.5 atm

A = 100 cm2

wqU

JwwU 25.2270

dTCndTCU mVV , mVCnUT ,

KmolJKmolJ 26.58.285.125.227 11

• n = 1.5 mol CO2

• CV,m = 28.8 JK-1mol-1

• A perfect gas• Adiabatic

e) ΔS=?From part d)

Question 5

15 cm

V1 V2

T1 = 15°C

P1 = 9 atm

Pext = 1.5 atm

A = 100 cm2

KT 26.5The gas undergoes constant-volume

cooling followed by isothermal expansion21 SSS

3

2 2

32

1 1

2 lnlnV

VnRdT

T

C

V

VnRdT

T

C VV

2

32

1

lnV

VnRdT

T

CV

• n = 1.5 mol CO2

• CV,m = 28.8 JK-1mol-1

• A perfect gas• Adiabatic

e) ΔS=?

Question 5

15 cm

V1 V2

T1 = 15°C

P1 = 9 atm

Pext = 1.5 atm

A = 100 cm2

TTT 12

2

32

1

lnV

VnRdT

T

CS V

2

3

1

2 lnlnV

VnR

T

TCV

KK 89.28226.5)15.27315(

21 VV Constant volume cooling

1

11 P

nRTV

atm

KmolKatmLmol

9

15.28810206.85.1 112

L941.3Find this first!

• n = 1.5 mol CO2

• CV,m = 28.8 JK-1mol-1

• A perfect gas• Adiabatic

e) ΔS=?

Question 5

15 cm

V1 V2

T1 = 15°C

P1 = 9 atm

Pext = 1.5 atm

A = 100 cm2

2

32

1

lnV

VnRdT

T

CS V

2

3

1

2 lnlnV

VnR

T

TCV

VVV 23

Lcm

LcmcmL 44.5

1

1015100941.3

3

32

L

LmolJKmol

K

KmolJKmol

941.3

44.5ln314.85.1

15.288

89.282ln8.285.1 1111

1224.301996.479588.0 JK