relation between c p & c v recall: can be related

Relation between CP & CV

VPVP T

U

T

HCC

VP

VP T

U

T

PVUCC

PVUH Recall:

VPPVP T

U

T

VP

T

UCC

Can be related

Relation between CP & CV

VPPVP T

U

T

VP

T

UCC

VT

U

U is a function of T and V

dVV

UdT

T

UdU

TV

Thus;

At constant P;

PT

PV

P dVV

UdT

T

UdU

Relation between CP & CV

PT

PV

P dVV

UdT

T

UdU

Dividing by dTP ;

P

P

TVP

P

dT

dV

V

U

T

U

dT

dU

PTVP T

V

V

U

T

U

T

U

VPPVP T

U

T

VP

T

UCC

VPPVP T

U

T

VP

T

UCC

PT

VP T

VP

V

UCC

Joule Experiment

 - A sample of a gas (the system) was placed in one side of the apparatus and the other side of the apparatus was evacuated.

-The initial temperature of the apparatus was measured.

-The stopcock was then opened and the gas expanded irreversibly into the vacuum. 

To determine TV

U

Joule Experiment

-Because the surroundings were not affected during the expansion into a vacuum, w was equal to zero.

-The gas expanded rapidly so there was little opportunity for heat to be transferred to or from the surroundings.

- If a change in temperature of the gas occurred, heat would be transferred to or from the surroundings after the expansion was complete, and the final temperature of the surroundings would differ from the initial temperature.

Joule Experiment

UJ VT /Joule coefficient, J

To determine TV

U

UVT V

T

T

U

V

U

JVT

CV

U

Joule experiment gave and 0J 0/ TVUbecause the changes in temperature that occurred were too small to be measured by the thermometer.

Joule-Thomson experiment

Initial State

Final State

Insulated wall

To determine TP

H

Joule-Thomson experiment

rightleft www

2

1 0

2

0

1

V

V

dVPdVP

2211 VPVP

wqUUU 12

w0Adiabatic process

221112 VPVPUU Rearranging;

111222 VPUVPU

12 HH

0HIsenthalpic

process

Joule-Thomson experiment Joule-Thomson coefficient,

HJT PT /

HPT P

T

T

H

P

H

JTPT

CP

H

To determine TP

H

JT

Perfect Gases Perfect gas: one that obeys both of the

following:nRTPV

0

TV

U U is not change with V at constant T

If we change the volume of an ideal gas (at constant T),

we change the distance between the molecules.

Since intermolecular forces are zero, this distance

change will not affect the internal energy.

For perfect gas, internal energy can be expressed as a function of temperature (depends only on T):

An infinesimal change of internal energy,

Perfect Gases

TUU

dTCdU VV

V dT

UC

From

(slide 19)

Perfect Gases For perfect gas, enthalpy depends only on

T;

Thus;

An infinesimal change of enthalpy,

PVUH

nRTUH

This shows that H depends only on T for a perfect gas

THH

From

(slide19)

dTCdH PP

P dT

HC

Perfect Gases The relation of CP and CV for perfect gas;

PTVP T

VP

V

UCC

From slide 24

PP T

VP

T

VP

0

Perfect gas

nRCC VP

PV=nRT, thus PnRT

V

P

/

RCC mVmP ,,or

Perfect Gases For perfect gas,

JTPT

CP

H

JVT

CV

U

0

TV

U

0

TP

H

- Since U & H depend only on T.

Perfect gas and First LawFirst law;

For perfect gas;

dwdqdU

dwdqdU

PdVdqdTCV Perfect gas, rev. process, P-V work only

Reversible Isothermal Process in a Perfect Gas First law:

Since the process is isothermal, ∆T=0; Thus, ∆U = 0

First law becomes;

wqU

dTCdU V

0

0

wq 0

Reversible Isothermal Process in a Perfect Gas Since , thus:

Work done;

qw 0wq

2

1

PdVw

2

1

2

1

1dV

VnRTdV

V

nRT

2

1lnV

VnRT

1

2lnP

PnRTOR rev. isothermal

process, perfect gas

Since

Reversible Adiabatic Process in Perfect Gas First law,

Since the process is adiabatic,

For perfect gas, becomes,

dwdqdU

0dq

dwdU

dwdU dWdTCV dTCdU VPdV

dVV

nRT

Reversible Adiabatic Process in Perfect Gas

Since , thus

By integration;

dVV

nRTdTCV

dVV

RTdTC mV

,

2

1

2

1

, dVV

RdT

T

C mV

2

1lnV

VR

Reversible Adiabatic Process in Perfect Gas

If is constant, becomes;

OR

mVC ,

2

12

1

, lnV

VRdT

T

C mV

2

1

,1

2 lnlnV

V

C

R

T

T

mV

mVC

R

V

V ,

2

1ln

mVC

R

V

V

T

T ,

2

1

1

2

rev. adiabatic

process, perfect gas

Alternative equation can be written instead of

Consider

Reversible Adiabatic Process in Perfect Gas

mVC

R

V

V

T

T ,

2

1

1

2

222111 // TVPTVP

121122 /TTVPVP

mVC

R

V

VVPVP

,

2

11122

mVmV CRCR VPVP ,, 122

111

Since

Reversible Adiabatic Process in Perfect Gas

mVmV CRCR VPVP ,, 122

111

mVmVmV CRCCR ,,,1

RCC mVmP ,,mVmP CC ,,2211 VPVP mVmP CC ,,where

rev. adiabatic process, perfect gas, CV constant

Reversible Adiabatic Process in Perfect Gas If is constant, becomesVC wqU

wTTCV 12rev. adiabatic process, perfect gas, CV constant

Exercise A cylinder fitted with a frictionless piston

contains 3.00 mol of He gas at P = 1.00 atm and is in a large constant- temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q and ∆ U for the process.

Solution A cylinder fitted with a frictionless piston

contains 3.00 mol of He gas at P = 1.00 atm and is in a large constant- temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q and ∆ U for the process.

Assumption: He as a perfect gas. ∆U =0 since T is constant (U depends only on T)

1

2lnP

PnRT

0 wqU

2

1

PdVw 2

1

2

1

1dV

VnRTdV

V

nRT

2

1lnV

VnRT

Solution

1

2lnP

PnRTw

00.100.5ln400134.800.3 11 KKJmolmol

J41061.1

Jwq 41061.1

1. Reversible phase change at constant P & T. Phase change: a process in which at least one new

phase appear in a system without the occurance of a chemical reaction (eg: melting of ice to liquid water, freezing of ice from an aqueous solution.)

Calculation of First Law Quantities

PqH

wqU

VPPdVw 2

1

Calculation of First Law Quantities2. Constant pressure heating with no phase

change

VPPdVw 2

1

dTCqHT

T

PP 2

1

wqwqU P

3. Constant volume heating with no phase change

Recall:

Calculation of First Law Quantities

02

1

PdVw

2

1

dTCqU v

dT

dqC vv

qwqU

4. Perfect gas change of state:

H & U of a perfect gas depends on T only;

Calculation of First Law Quantities

dTCdH P dTCdU V

dTCH P2

1

dTCU V2

1

5. Reversible isothermal process in perfect gas H & U of a perfect gas depends on T only;

Since , thus

Calculation of First Law Quantities

dTCH P2

1

dTCU V2

1

0H 0U

2

12

1

lnV

VnRTPdVw

0 wqU wq

6. Reversible adiabatic process in perfect gas

If Cv is constant, the final state of the gas can be found from:

where

Calculation of First Law Quantities

0q Adiabatic process

dTCH P2

1

dTCU V2

1

wwqU

2211 VPVP

VP CC

Exercise CP,m of a certain substance in the temperature

range 250 to 500 K at 1 bar pressure is given by CP,m = b+kT, where b and k are certain known constants. If n moles of this substance is heated from T1 to T2 at 1 bar (where T1 & T2 are in the range of 250 to 500 K), find the expression for ΔH.

SOLUTION CP,m of a certain substance in the temperature range 250 to

500 K at 1 bar pressure is given by CP,m = b+kT, where b and k are certain known constants. If n moles of this substance is heated from T1 to T2 at 1 bar (where T1 & T2 are in the range of 250 to 500 K), find the expression for ΔH.

- P is constant throughout the heating, thus;

dTCqHT

T

PP 2

1

dTnCT

T

mP2

1

,dTkTbn

T

T 2

1

mPC ,

2

12

2 T

T

kTbTn

22

12 122

1TTkTTbn