Review , Examples and

Problems

Thermodynamics221PHYS

HeatHeat

Example-1: An m1 = 485-gram brass block sits in boiling water (T1 = 100° C). It is taken out of the boiling water and placed in a cup containing m2 = 485 grams of ice water (T2 = 0° C). What is the final temperature, TF, of the system (i.e., when the two objects have the same T)? (cbrass = 380 J/kg.K; cwater = 4184 J/kg.K)

a. TF < 50° C b. TF = 50° C c. TF > 50° C

Solution:Heat flows from the brass to the water. No work is done, and we assume that no energy is lost to the environment.

Remember: Q = C∆T = mc∆TBrass (heat flows out): Q1 = ∆U1 = m1c1(TF-T1) Water (heat flows in): Q2 = ∆U2 = m2c2(TF-T2)

Energy is conserved: Q1 + Q2 = 0Solve for TF: TF = (m1c1T1+m2c2T2) / (m1c1+m2c2)

= (c1T1+ c2T2) / (c1+ c2) = 8.3° CWe measured TF = _____° C.

QT1

m1

T2

m2

Kinetic theory Kinetic theory of the ideal of the ideal

gasgas

Example2 : The Speed of Molecules in AirAir is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0 u) and oxygen O2 molecules (molecular mass 32.0 u). Assume that each behaves as an ideal gas and determine the rms speeds of the nitrogen and oxygen molecules when the temperature of the air is 293K.

2 312 2

3rms rms

kTmv kT v

m= ⇒ =Q

Solution

For Nitrogen…

kg1065.4g1065.4mol106.022

molg0.28 2623123

−−− ×=×=

×=m

( )( )sm511

kg1065.4

K293KJ1038.13326

23

=×

×== −

−

m

kTvrms

Example 3: A container of an ideal gas has a moveable top. The top has an area of 0.01 m2 and is 50 cm above the bottom of the cylinder. A mass of 200 kg is placed on the container, which compresses the gas by 20 cm. The gas in the container is initially at atmospheric pressure (1.01 x 105 Pa) and 20oC. What is the new temperature of the gas?

nRTPV = nRT

PV =2

22

1

11

T

VP

T

VP =Solution

1 2

1 1 2 2

V Ah and A A

V Ah and V Ah

= =⇒ = =

Q

15.27311 += cTT K15.293=

1 0 101P P kPa= =

A

FPP += 02

A

mgPP += 02 kPa2.297=

The following relations have been used

11

1222 VP

TVPT = K6.517=

10

122

AhP

TAhP=10

122

hP

ThP= CT 2442 =

The new temperature

Thermodynamic Thermodynamic lawslaws

Ex.4 : The process shown on the Pressure-Volume diagram is an

(A) adiabatic expansion.

(B) isothermal expansion.

(C) isometric expansion.

(D) isobaric expansion.

P

Vo

Ex.5: In an isochoric process, there is no change in

(A) pressure.

(B) temperature.

(C) volume.

(D) internal energy.

Ex-6: The process shown on the Temperature-Volume graph is an

(A) adiabatic compression.

(B) isothermal compression.

(C) isochoric compression.

(D) isobaric compression.

T

Vo

Ex-7: When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an isothermal process,

(A) ΔU = 0

(B) W = 0

(C) Q = 0

(D) none of the above

Ex-8: An ideal gas is compressed to one-half its original volume during an isothermal process. The final pressure of the gas

(A) increases to twice its original value.

(B) increases to less than twice its original value.

(C) increases to more than twice its original value.

(D) does not change.

Ex-10: When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an adiabatic process,

(A) ΔU = 0.

(B) W = 0.

(C) Q = 0.

(D) none of the above

Ex-11: A gas is taken through the cycle illustrated here. During one cycle, how much work is done by an engine operating on this cycle?

(A) PV

(B) 2PV

(C) 3PV

(D) 4PV

P

2P

V 2V 3V 4V

Ex-12: An ideal gas initially has pressure Po, at volume Vo and absolute temperature To. It then undergoes the following series of processes:

Po

2Po

Vo 2Vo 3Vo

I. Heated, at constant volume to pressure 2Po

II. Heated, at constant pressure to pressure 3Vo

III. Cooled, at constant volume to pressure Po

IV. Cooled, at constant pressure to volume Vo

I

II

III

IV

Po

2Po

Vo 2Vo 3Vo

I

II

III

IV

•

••

•2To

Find the temperature at each end point in terms of To

nRTPV =nR

VPT oo

o =

To

6To

3To

Po

2Po

Vo 2Vo 3Vo

I

II

III

IV

•

••

•Find the net work done by the gas in terms of Po and Vo

ooVP2W =

Net work equals net area under curve

Po

2Po

Vo 2Vo 3Vo

I

II

III

IV

•

••

•

Find the net change in internal energy in terms of Po and Vo

0U =∆

Ex-13: A sample of gas expands from 1.0 m3 to 4.0 m3 while its pressure decreases from 40 Pa to 10 Pa. How much work is done by the gas if its pressure changes with volume via each of the three paths shown in the Figure below?

Path A: W = +120 JPath B: W =+75 JPath C: W =+30 J

Example-14Example-14

A cylinder of radius 5 cm is kept at pressure with a pistonof mass 75 kg.

a) What is the pressure inside the cylinder?b) If the gas expands such that the cylinder rises 12.0 cm, what work was done by the gas?c) What amount of the work went into changing the gravitational PE of the piston?d) Where did the rest of the work go?

SolutionGiven: M =75, A = π× 0.052, ∆x=0.12, Patm = 1.013x105 Pa

a) Find Pgasatmgas P

A

MgP += = 1.950x105 Pa

b) Find Wgas

VPW ∆=

xAPW ∆= gas = 183.8 J

c) Find Wgravity mghW = = 88.3 J

d) Where did the other work go? Compressing the outside air

Example-15 a) What amount of work is performed by the gas in the cycle IAFI?

b) How much heat was inserted into the gas in the cycle IAFI?

c) What amount of work is performed by the gas in the cycle IBFI?

area enclosed=WWIAFI = 3Patm

= 3.04x105 J

WQU −=∆∆U = 0

W = -3.04x105 J

Q = 3.04x105 J

V (m3)

Example-Example-1616Consider a monotonic ideal gas which expands according to the PV diagram.

a) What work was done by the gas from A to B?b) What heat was added to the gas between A and B?c) What work was done by the gas from B to C?d) What heat was added to the gas beween B and C?e) What work was done by the gas from C to A?f) What heat was added to the gas from C to A?

V (m3)

P (kPa)

25

50

75

0.2 0.4 0.6

A

B

C

Solutiona) Find WAB

b) Find QAB

V (m3)

P (kPa)

25

50

75

0.2 0.4 0.6

A

B

C

WAB = Area = 20,000 J

•First find UA and UB

Monotonic Gas:

3

2

3

2

U nRT

PV nRT

U PV

=

=

=

UA = 22,500 J, UB = 22,500 J, ∆U = 0

•Finally, solve for QU Q W∆ = − Q = 20,000 J

c) Find WBC

d) Find QBC

V (m3)

P (kPa)

25

50

75

0.2 0.4 0.6

A

B

C

WBC = -Area = -10,000 J

•First find UB and UC

PVU2

3=

UB = 22,500 J, UC = 7,500 J, ∆U = -15,000

•Finally, solve for Q

WQU −=∆

Q = -25,000 J

e) Find WCA

f) Find QCA

V (m3)

P (kPa)

25

50

75

0.2 0.4 0.6

A

B

C

WAB = Area = 0 J

•First find UC and UA

UC = 7,500 J, UA = 22,500 J, ∆U = 15,000

•Finally, solve for Q

Q = 15,000 J

g) Net work done by gas in the cycleh) Amount of heat added to gas

WAB + WBC + WCA = 10,000 JQAB + QBC + QCA = 10,000 J

This does NOT mean that the engine is 100% efficient!

Example-17

Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature?

γγ2211

222

111

VPVP

TNkVP

TNkVP

B

B

=

==

1

2

1

2

12

1

1121

2

11

2

112 T

T

V

VT

T

VPTNk

V

VP

V

VPP B =

⇒===

−

−

γ

γ

γ

γ

γ

constVTVT == −− 122

111

γγFor adiabatic processes:

Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity

constTP =− γγ /1also

KKKV

VTT 51474.12954295 4.0

1

2

112 ≈×≈×=

=

−γ

Example-18: Finding the Work

An ideal gas with γ = 1.4 occupies 4.0 L at 300 K & 100 kPa pressure. It’s compressed

adiabatically to ¼ of original volume, then cooled at constant V back to 300 K, & finally

allowed to expand isothermally to its original V. How much work is done on the gas?

1A A B B

AB

p V p VW

γ−=−

741 J= −

AB (adiabatic):

0BCW =BC isochoric):

ln ACA

C

VW n R T

V=CA

(isothermal):

( ) ( ) ( )1.4 1100 4.0 1 4

1.4 1

kPa L −−=

−

AB A

B

Vp p

V

γ

= ÷

1

11

A A AAB

B

p V VW

V

γ

γ

− ÷= − ÷ ÷−

ln 4A Ap V= 555 J=

work done by gas: ABCA AB BC CAW W W W= + + 186 J= −

Example-19: A thermodynamic system undergoes a process in which its internal energy decreases by 465 J. Over the same time interval, 236 J of work is done on the system. Find the energy transferred from it by heat.

465 236 701

U Q W

Q U W J J J

∆ = += ∆ − = − − = −

Note: Sign convention for Q : Q>0 system gains heat from environment

Example-20:. Diesel Power

Fuel ignites in a diesel engine from the heat of compression (no spark plug needed).

Compression is fast enough to be adiabatic.

If the ignit temperature is 500°C, what compression ratio Vmax / Vmin is needed?

Air’s specific heat ratio is γ = 1.4, & before the compression the air is at 20 °C.

1T V constγ − =

( )1 / 1.4 1273 500

273 20

K K

K K

− += ÷+

( )1 / 1

max min

min max

V T

V T

γ −

= ÷

11=

Ex-21: Two cylinders at the same temperature contain the

same gas. If B has twice the volume and half the number of

moles as A, how does the pressure in B compare with the

pressure in A?(A) PB = 1/2 PA

(B) PB = 2 PA

(C) PB = 1/4 PA

(D) PB = 4 PA

(E) PB = PA

Ex-22: A gas cylinder and piston are covered with heavy insulation. The piston is pushed into the cylinder, compressing the gas. In this process, the gas temperature

A. doesn’t change.B. decreases.C. increases.D. there’s not sufficient information to tell.

Ex-23: During an isothermal process, 5.0 J of heat is removed from an ideal gas. What is the change in internal energy?

A) zero

B) 2.5 J

C) 5.0 J

D) 10 J

The Laws of Thermodynamics

Ex-24: If the gas in a container absorbs 300 J of heat, has 100 J of work done on it, and then does 200 J of work on its surroundings, what is the increase in the internal energy of the gas?

(A) 600 J(B) 400 J(C) 0 J(D) 500 J(E) 200 J