Physics 2101 Physics 2101 Section 3Section 3Section 3Section 3

May 3May 3rdrd: Chap. 19: Chap. 19

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Pressure, Temperature, & RMS speed

h ll f h l l h h llAssume the collision of the gas molecule with the wall is elastic then:

( ) ( ) 2x x x xp mv mv mvΔ = − − = −

The molecule travels to the back wall, collides and comes back.  The time it takes is 2L/vx.

( ) ( )x x x xp

222 /

x x

x

mv mvpt L v L

Δ= =

ΔBut the pressure is F/A2 If we calculated the average velocity and(v2 )

p =Fx

A=

1L2

mvxi2

Li=1

n

∑ =mL3 vxi

2

i=1

n

∑If we calculated the average velocity          and use the fact that the number in the sum is nNA then: 

(vx )avg

2

p =nmNA

L3 (vx2 )avg =

nM (vx2 )avg

VnM (v2 ) M 2

v2 = vx2 + vy

2 + vz2

2 v2(v2 )avg = vrms

=nM (v )avg

3V=

nMvrms2

3Vvx

2 =3 RMS = Root‐Mean‐Square

M ‐‐‐Molar mass

RMS Speeds

p =nM v 2( )ave

3VWe have

pV = nRTFor ideal gas

1/ 2 1/ 23 3rms

pV nRTv ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 (19-22)

rms

rms

nM nM

RTvM

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=M

The RMS velocity depends on:

Molar mass & Temperature

Problem 19-3: Here are five numbers: 5, 11, 32, 67, and 89. (a) What is the average value navg?(b) Wh t i th l f th b ?(b) What is the rms value nrms of the numbers?

5 + 11 + 32 + 67 + 89

(a)

navg =5 + 11 + 32 + 67 + 89

5= 40.8

1 n

∑ 52 + 112 + 322 + 672 + 892

(b)

(n2 )avg =1n

ni2

i=1∑ =

5 + 11 + 32 + 67 + 895

= 2714.41

(n2 ) = 52 1(n )avg = 52.1

Problem 19-18: Calculate the rms speed of helium atoms at 1000K. Helium has 2 protons and 2 neutrons, what is the molar mass?

3Appendix F: 4.00 10 /U E 19 22

M x kg mol−=Use Eqn 19-22

3rms

RTv =rmsvM

( )( ) 33 8,31 / 10003 J molK KRT ( )( ) 33

3 8,31 / 10003 2.50 10 /4.00 10 /rms

J molK KRTv x m sM kg mol−= = =

×

2mv3kTK

Translational Kinetic Energy

22rms

The kinetic energy of a gas molecule .2

Its average kinetic energy .

mvK

mvmvK

=

⎛ ⎞= =⎜ ⎟

avg 2K =

avgavg

avgA

Its average kinetic energy .2 2

3 3Thus .2 2

K

m RT RTKM N

⎜ ⎟⎝ ⎠

= =

av

A

g

2 2

We finally get: 3 2

K

MkT

N

=

At a given temperature alT l ideal gas molecules, no matter what their mass, have the same average translational kinetic energy. When we measure the temperature of a gas, we are also measuring the average translationalkinetic energy of its molecules.

Problem 19-26: What is the average translational kinetic energy of nitrogen molecules at 1600 K?

Average Kinetic energy is

Kavg =32

kT

Kavg =32

kT =32

1.38x10−23 J( )1600K( )= 3.31x10−20 J

Molar Specific Heat: Monatomic Ideal GasMolar Specific Heat at Constant Volume (Wb =0)Molar Specific Heat at Constant Volume (Wby 0)

Q = nCV ΔT & Wby = 0

ΔE int = Q = nCV ΔT Always true ifV = constV = const.

ΔE int = n 32 R( )ΔT = n CV( )ΔT = Q ( )

CV ,monatomic =32

R = 12.5J/mol ⋅ K

Molar Specific Heat: Monatomic ideal gas

l f l ( )

ΔE int = Q = nCV ΔT

Molar Specific Heat at Constant Volume (Wby=0)

Always true for const. VV

CV ,monatomic =32

R = 12.5 ⋅ J/mol ⋅ K

y

Molar Specific Heat at Constant Pressure (Wby=pΔV)

ΔE int = Q − W = nCpΔT − pΔVnCV ΔT = nCpΔT − nRΔT

CV = C p − R or C p = CV + R

Cp,monatomic =52

R = 20.8 ⋅ J/mol ⋅ K

Molecular Specific heat Constant Volume CV

CV ,monatomic =32

R = 12.5 ⋅ J/mol ⋅ K

ΔEi t = nCV ΔTΔEint nCV ΔT

Work Done by Isothermal (ΔT = 0)Expansion/Compression of Ideal Gas

On p‐V graph, the green lines are isotherms…… each green line corresponds to a system at a constant temperature.

From ideal gas law this means that for a given isotherm:

Relates p and V

From ideal gas law, this means that for a given isotherm:

pV = constant ⇒ p = nRT( ) 1V

The work done by the gas is then: 

V ⎛ ⎞V V

⇒ Wby,isothermalΔT =0

= nRT lnVf

Vi

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛ ⎞Wby = pdV

Vi

V f

∫ =nRTV

⎛ ⎝ ⎜

⎞ ⎠ ⎟ dV = nRT

Vi

V f

∫ dVVVi

V f

∫ = nRT ln pi

p f

⎛

⎝ ⎜

⎞

⎠ ⎟

Adiabatic Expansion of an ideal gasBecause a gas is thermally insulated, or expansion/compression happens suddenly  ⇒ adiabatic

Remember “Adiabatic means Q = 0”

or, by 1st Law of Thermo ⇒ ΔEint = ‐Wby

In this case           pVγ = constant where     γ = Cp/CV = (R+ CV )/CV

example: monatomic gas γ = 5/3

p V γ = p V γp1V1 = p2V2

T1V1γ −1 = T2V2

γ −1

Compare with Isothermal Expansion (ΔT = 0)Compare with Isothermal Expansion (ΔT   0)

T1 = T2 ⇔ p1V1 = p2V2[ ]isothermal

Adiabatic Expansion of an ideal gasRemember “Adiabatic means Q = 0”

or, by 1st Law of Thermo ⇒ ΔEint = ‐WbypVγ = constant

P fProof

dEint = Q − pdVdE = − pdV = nC dt

−p

CV

⎛⎝⎜

⎞⎠⎟

dV =PdV + Vdp

Cp − CVdEint = pdV = nCV dt

ndT = −p

CV

⎛⎝⎜

⎞⎠⎟

dV

p

dpp

+Cp

CV

⎛

⎝⎜⎞

⎠⎟dVV

= 0

pV = nRTor pdV+Vdp=nRdT

dpp

+ γdVV

= 0

Remember Cp − CV = Rgiving

Integrating givesln p + γ lnV = Constant

ndT =PdV + Vdp

Cp − CV

p γ

or pVγ = Constant

Consider the ideal gas in fig. . The container is welli l t d Wh th d h t i t f d

aAdiabatic Expansion of an Ideal Gas

i i f fpV p Vγ γ=1 1

i i f fTV T Vγ γ− −= insulated. When the gas expands, no heat is transferred to or from the gas. This process is called adiabatic. Such a process is indicated on the - diagram of fig. p V b

i i f fTV T V

by the red line. The gas starts at an initial pressure andinitial volume . The corresponding final parametersare and The proce

i

i

pV

p V ss is described by the equationare and . The procef fp V ss is described by the equation

Here the constant ..i i f fp

V

CC

pV p Vγ γ γ ==

1

Using the ideal gas law we can get the equation

V γ −1 1

1

1

If we have adiaba

i i f fi

f if

f i

VT TV

V

TV T V

V

γγ

γγ

−−

− → =

>

=

tic expansion and .f iT T<f

If we have adiabatic compression and .f

f i f iV V T T< >

i i f fpV p V=i fT T=

Free ExpansionIn a free expansion, a gas of initial volume Vi and initial

i ll d t d i t t ipressure pi is allowed to expand in an empty containerso that the final volume is Vf and the final pressure p f .

In a free expansion 0 because the gas container is insulated. Furthermore,since the expansion takes place in vacuum the net work 0.

QW

==

intThe first law of thermodynamics predicts that 0.Since the g

EΔ =as is assumed to be ideal there is no change in temperature: T T=Since the gas is assumed to be ideal there is no change in temperature: .

Using the law of ideal gases we get the following equation,

i fT T=

which connects the initial with the final state of the gas:.i i f fpV p V=

int3

2nRTE =

Internal Energy of an Ideal GasConsider a monatomic gas such as He, Ar, or Kr. In this case the internal energyEint of the gas is the sum of the translational kinetic energies of the constituentatoms. The average translational kinetic energy of a single atom is given by the

i K3kT

A l f l i N Nequation Kavg =2

. A gas sample of n moles contains N = nNA atoms.

The internal energy of the gas Eint = NKavg =nNA 3kT

2=

3nRT2

.g 2 2

The equation above expresses the following important result:

The internal energy Eint of an ideal gas is a function of gas temperature only; it does not depend on any other parameter. only; it does not depend on any other parameter.

Work Done by Isobaric (ΔP = 0) Expansion of an Ideal Gasp

⇒ Wby,isobaricΔP=0

= pΔV

=nRΔT

Problem 19-6: A quantity of ideal gas at 10.0 0C and 100 kPaoccupies a volume of 2.50 m3. (a) How many moles of the gas

t? (b) If th i i d t 300 kP dare present? (b) If the pressure is now raised to 300 kPa and the temperature to 30.0 0C, how much volume does the gas occupy?

pV 100x103 Pa( ) 2.50m3( )106 ln =

pRT

=( )( )8.31J / mol • K( ) 283K( )

= 106moles

(b) pfVf

piVi

=Tf

Tipi i i

Vf = Vipi

p f

⎛

⎝⎜

⎞

⎠⎟

Tf

Ti

⎛

⎝⎜⎞

⎠⎟= 0.892m3

pf⎝ ⎠ i

Remember special cases…Ideal Gas:Ideal Gas: ( )3E nR TΔ = ΔPV nRT=

Adiabatic expansion/contraction ‐ NO TRANSFER OF ENERGY AS HEAT    Q = 0

Ideal Gas: Ideal Gas: ( )int 2E nR TΔ = ΔPV nRT=

ΔE int = −W[ ]adiabatic

Wby = pdV∫= area under p − V graph

Constant‐volume processes (isochoric)‐NOWORK IS DONE W = 0 W = pdV

V f =Vi

∫ = 0 area under p V graph

ΔE int = QNO WORK IS DONE     W = 0 Wby = pdV

Vi

∫ = 0

QΔV = 0 = nCV ΔT

Wby = pdVVi

V f =Vi

∫ = pΔV

QΔP 0 = nCPΔT

Constant‐pressure processes

ΔE int = Q − pΔVCV = CP − R

3) Cyclical process (closed cycle) ΔEint,closed cycle =0a) net area in p‐V curve is Q ΔE int = 0 ⇒ Q = W

QΔP= 0 CP

19#46: One mole of an ideal atomic gas goes from a to c along the diagonal pate. The scale of the g g pvertical axis is set by pab=5.0  kPa and pc=2.0 kPa, and the scale of the V axis is Vbc4.0 m3 and Va=2.0 m3.  (a) what is the change in the internal Energy?  (b) How much Energy is added to the gas?  (c) How much heat is required if the gas goes from a to c via abc?

For any straight line on pv plotit is easy to prove

⎛ ⎞

Eint = n32

⎛⎝⎜

⎞⎠⎟

RT

Wstraight =pf + pi

2⎛⎝⎜

⎞⎠⎟

ΔV

3⎛ ⎞ 3⎛ ⎞(a) Eint c − Eint a =32

⎛⎝⎜

⎞⎠⎟

Tc − Ta( )=32

⎛⎝⎜

⎞⎠⎟

pcVc − paVa( )= −3000J

(b) W =pa + pc⎛

⎝⎜⎞⎠⎟

V − V( )= 7000J (c) ΔEint found in (a)(b) W2⎝⎜ ⎠⎟

Vc Va( ) 7000J

Q = ΔEint + W = 2000JW = 5.0x103 Pa( ) 2m3( )= 10000J

Q = 5000J

19 #63: 1.00 mol of an ideal monatomic gas goes through the cycle shown in the Figure. The temperatures are T1the cycle shown in the Figure. The temperatures are T1=300 K, T2=600 K, and T3= 455K. For 1 to 2, what are (a) heat Q, (b) the change in internal energy, and (c) the work done W?  For 2 to 3, what are (d) Q, (e) change in Eint, and (i) W? F th f ll l h t (j) Q (k) h i E(i) W? For the full cycle, what are (j) Q, (k) change in Eint, and (l) W.  The Initial pressure at point 1 is 1.00 atm. What are the (m) volume and (n) pressure at point 2 and the (o) volume and (p) pressure at point 3.(p) p p

( ) ( )3int 2 12(a, b, c) For 1 2, 0, 0, ; Here VV W E n R T n C T Q T T T⇒ Δ = = Δ = Δ = Δ = Δ = −

( )3int 3 22(d, e, f) For 2 3, Q 0, ; Here E n R T W T T T⇒ = Δ = Δ = − Δ = −

( )3For 3 1 0; 0;E n R T W p V nR T⇒ Δ = Δ < = Δ = Δ <( )int 2

int 1 3

For 3 1, 0; 0;

5 ; Here 2

E n R T W p V nR T

Q E W nR T T T T

⇒ Δ = Δ < = Δ = Δ <

= Δ + = Δ Δ = −

Problem 19-21: (a) Compute the rms speed of a nitrogen molecule at 20.0 0C. Each N atom has 7 protons and 7 neutrons? (a) what is the rmsspeed at 300K and 20 0 0C At what temperatures will the rms speed bespeed at 300K and 20.0 C. At what temperatures will the rms speed be (b) half that value and (c) twice that value?

Table 19-1: 28 /M g mol=(a) Use Eqn 19-22

3RTv =Set up ratios

3 /v RT M T

517 / (for 300 )

rms

rms

vM

v m s T K= =

2 2 2

1 1 1

3 /3 /

v RT M Tv RT M T

= =

1 517 / for = 300K

293

v m s T

K

= 23( ) Set v and solve

2vb =

( )2

2

293 517 /300

511 /

Kv m sK

v m s

=

=

3 73T K=

4 2

4

( ) Set v 2 and solve 1170c v

T K=

=