subject chemistry paper no and title 10, physical chemistry...

CHEMISTRY

Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics)

Module No. 4: Isothermal and Adiabatic expansion

Subject Chemistry

Paper No and Title 10, Physical Chemistry- III (Classical Thermodynamics,

Non-Equilibrium Thermodynamics, Surface chemistry, Fast

kinetics)

Module No and Title 4, Isothermal and Adiabatic expansion

Module Tag CHE_P10_M4

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TABLE OF CONTENTS

1. Learning Outcomes

2. Expansion of ideal gas

3. Isothermal expansion

3.1 Work done in reversible isothermal expansion

3.2 Work done in Reversible Isothermal Compression

3.3 Work done in Irreversible Isothermal Expansion

4. Adiabatic expansion

4.1 Final temperatures in reversible and irreversible adiabatic expansion

4.1.1. Reversible adiabatic expansion

4.1.2. Irreversible adiabatic expansion

5. Reversible isothermal expansion of real gas

5.1 Work of expansion, w

5.2 Internal energy change, U

5.3 Enthalpy change, H

5.4 Heat change, q

6. Comparison of work of expansion of an ideal gas and van der Waal gas

7. Summary

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1. Learning Outcomes

After studying this module you shall be able to:

Know about the effect of expansion on the thermodynamic properties like work done (w),

change in internal energy (ΔU), enthalpy change (ΔH) etc.

Differentiate between the work done in isothermal expansion and isothermal compression.

Know about the dependency of thermodynamic quantities q, w, H on temperature in the

case of adiabatic expansion.

Differentiate the change in enthalpy in the case of real gas and ideal gas.

2. Expansion of ideal gas

From the first law of thermodynamics we can calculate the possible changes in

thermodynamic properties such as q, w, ΔU and ΔH when expansion occurs in ideal gas. The

expansion may be isothermal or adiabatic and expansion can be reversible or irreversible.

3. Isothermal expansion

Calculation of ΔU

In isothermal expansion, the temperature of the system does not vary. The internal energy U

of an ideal gas is the function of temperature, so at constant temperature (isothermal process)

the internal energy of the system remains constant. Therefore, ΔU=0.

Calculation of ΔH

Thermodynamically, H=U+PV …(1)

Therefore, ΔH=Δ(U+PV)

= ΔU+ΔPV = ΔU+Δ n RT (from ideal gas equation) …(2)

But in isothermal process, ΔT=0 and ΔU=0, therefore ΔH=0

Calculation of q and w

From the first law of thermodynamics we derived that ΔU= q + w. Since in isothermal process

ΔU = 0, hence w= -q. This shows that by absorption of heat the work is done by the system in

isothermal expansion.

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3.1 Work done in reversible isothermal expansion

Consider a system in which gas is enclosed in a cylinder with frictionless and weightless

piston. The temperature of the system remains constant since system is supposed to be in

thermal equilibrium with the surroundings. The external pressure P on the piston is equal to the

pressure of the gas within the cylinder. The expansion of the gas occurs and the volume of the

gas changes from V to V+dV when the external pressure of the gas is lowered by infinitesimal

amount dP to become P-dP. Because of expansion, the pressure of the gas becomes equal to

the external pressure. The piston then comes to rest. Further if the process is repeated and the

external pressure is lowered by infinitesimal amount dP then the gas will undergo second

infinitesimal expansion dV before the pressure again equals to new external pressure.

Continuing the process, the external pressure will further be lowered which results in increase

of volume dV each time. Since the system is in thermal equilibrium with the surroundings, the

infinitesimally small cooling produced as a result of infinitesimally small expansion of the gas

at each step, is offset by the heat absorbed from the surroundings and the temperature remains

constant throughout the operation. During expansion, pressure decreases and the volume

increases. Thus the work done by the gas for the small volume change in an infinitesimal

expansion is given by :

dw = – (P – dP)dV = – PdV …(3)

Since the product dPdV is very small quantity therefore ignored. The total work done w by the

gas in expanding from volume V1 to the volume V2 will be the sum of the series of the terms

PdV in which pressure keeps on decreasing and is given by

V2

V1

w P dV …(4)

Substituting P = RT/V in the above expression for one mole of an ideal gas thus giving,

V2 2

V 11

VdVw RT RT ln

V V …(5)

At constant temperature, 1 1 2 2P V P V for the case of ideal gas, thus work done can also be

written as

1 2w RT ln (P / P ) …(6)

For n moles of the gas, the work done can be written as:

2 1 1 2w nRT ln (V / V ) nRT ln (P / P ) =−2.303 𝑛𝑅𝑇 𝑙𝑜𝑔𝑃1

𝑃2 …(7)

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As V2 is greater than V1 in case of expansion and P2 is less than P1 thus in above equation work

done will always be negative.

The expansion of a given amount (n mole) of gas results into decrease in the moles per unit

volume. Therefore equation (5) can be modified to represent the work done when an ideal

solution is diluted from initial concentration c1 to a final concentration c2.

Since 𝑐𝑖 =𝑛𝑖

𝑉 therefore, 𝑤 = −𝑛𝑅𝑇 𝑙𝑛

𝑐1

𝑐2

3.2 Work done in Reversible Isothermal Compression

Now suppose gas is compressed reversibly from volume V2 to V1. In this case, the external

pressure will be greater than pressure inside the cylinder by an infinitesimal amount dP. Thus

external pressure will be P+dP. Let the gas is compressed by an infinitesimal amount, say dV.

The two parameters pressure and volume are of opposite signs as during compression the

pressure increases and the volume decreases. Now the expression for the infinitesimal amount

of work done on the system by the surroundings is

dw (P dP) dV PdV …(8)

Here the term dPdV is small quantity therefore it is ignored.

Let the gas is compressed from volume V2 to V1, then the work done (w’) by the surroundings

on the gas is given by:

V1

V2

w P dV

Taking the gas to be ideal, thus substituting P= RT/ V and 1 1 2 2P V P V in the above expression

we get,

V1

1 2V2

dVw RT RT ln (V / V )

V = 2 1RT ln (P / P ) …(9)

For n moles of the gas, the work done is given by:

1 2 2 1w nRTln (V / V ) nRT ln (P / P ) …(10)

Since during compression the initial volume V2 > V1 and P2 < P1 thus w’ will always be

positive.

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3.3 Work done in Irreversible Isothermal Expansion

In this section two types of irreversible isothermal expansion will be discussed

These are: expansion against zero pressure i.e. in vacuum(called free expansion) and expansion

against constant external pressure which should be less than the pressure of the gas i.e. Pext <

P (called intermediate expansion)

Free expansion

In free expansion the external pressure is zero, therefore the work done is given by

extw dw P dV 0 …(11)

Intermediate expansion

Let the volume expands from V1 to V2 against an external pressure Pext then the work done will

be

V2 ext ext2 1

V1

w P dV P (V V ) …(12)

The work done during intermediate isothermal expansion is less than the work done during

reversible isothermal expansion since Pext is less than P in the case of intermediate isothermal

expansion while Pext is almost equal to P in the case of reversible isothermal expansion.

The work done by the gas during expansion is (-w),

Therefore, −𝑤 = 𝑃𝑒𝑥𝑡 (𝑛𝑅𝑇

𝑃2−

𝑛𝑅𝑇

𝑃1)

−𝑤 = 𝑛𝑅𝑇 (1 −𝑃2

𝑃1) (if P2 ≈ Pext)

4. Adiabatic expansion

In adiabatic expansion no heat is transferred between the system and the surroundings i.e. q=0

Thus equation for first law of thermodynamics becomes

U 0 w or w U …(13)

Since in expansion, work is done by the system so w will be negative. Thus U will also be

negative. Due to the decrease in internal energy the temperature of the system also decreases.

This states that work is done at the expense of internal energy of the gas. While in compression

w is positive therefore U will be positive which consequently increases the temperature of the

system. This states that the work is stored in the system in the form of rise in internal energy.

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Calculation of U .

At constant volume the molar heat capacity for an ideal gas is given by:

V VC ( U / T) …(14)

Accordingly finite change in internal energy will be

…(15)

Calculation of H

From classical thermodynamics, enthalpy is given by

H = U + PV

So, change in enthalpy for one mole of the gas is

H U (PV) U R T …(16)

Substituting the value of U , we get

V VH C T R T (C R) T = PC T …(17)

Calculation of w

In adiabatic process q=0, thus from the first law equation U = q + w, the work done will be

given by:

Vw U C T

Thus from the above equations we can say that all the three quantities U , H and w are

dependent on temperature. Since the variation in temperature depends upon the nature of the

process i.e. whether the process is reversible or irreversible, thus the magnitude of

thermodynamic properties will also vary with the nature of the process.

4.1 Final Temperatures in Reversible and Irreversible Adiabatic Expansions

4.1.1 Reversible Adiabatic Expansion. The final temperature in the case of reversible

adiabatic expansion is obtained from the expressions which relate the initial and final

temperatures to the respective volumes or pressures.

Relation between Temperature and Volume. Let the expansion is done against the external

pressure P and V be the increase in volume. Then, the external work done by the system is

equal to – P V . Hence, according to the First law equation,

U = – P V …(18)

VU C dT

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Let the fall in temperature be T and U = VC T

Therefore, VC T = – PdV ...(19)

For infinitesimally small quantities,

VC dT PdV RT dV / V (for 1 mole of the gas)

Dividing the above equation by T

VC dT / T R dV / V

Or VC d(ln T) R d(ln V)

Or 𝐶𝑣

𝑅 𝑑(ln 𝑇) = −𝑑(ln 𝑉) ...(20)

Integrating the above equation between the limits of temperatures T1 and T2 with the

corresponding volumes V1 and V2 then we get,

V 2 1 2 1 1 2C ln (T / T ) R ln (V / V ) R ln (V / V ) ...(21)

or 2 1 V 1 2ln (T / T ) (R / C ) ln (V / V )

Since, P VC C R and substituting P VC / C , we get

𝑙𝑛 (𝑇2

𝑇1) =

𝐶𝑣

𝑅= 𝑙𝑛 (

𝑉1

𝑉2) (for one mole)

𝑇2 = 𝑇1 (𝑉1

𝑉2)

𝑛𝑅

𝐶𝑣 = 𝑇1 (𝑉1

𝑉2)

𝑅

𝐶𝑣,𝑚

𝑇2 = 𝑇1 (𝑉1

𝑉2)

𝛾−1

…(22)

or 11 2 2 1T / T (V / V ) (Note that > 1)

Since in the case of expansion 2 1 2 1V V and T T . Hence, a gas cools during reversible

adiabatic expansion.

4.1.2 Irreversible Adiabatic Expansion.

In this section, two cases are considered:

Free Expansion.

In the case of free expansion the external pressure is zero, thus the work done in this expansion

is also zero. Therefore, U which is equal to w, is also zero. In the case of ideal gas, the change

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in the internal energy is function of temperature thus accordingly if U is zero then T is also

zero. The change in enthalpy is given by H = U + nR T , as U and T are zero therefore

H is also zero. Thus, in a free adiabatic expansion, T = 0, w = 0 and H = 0.

Intermediate Expansion.

Let the gas expands from the volume V1 to V2 against a constant pressure ExtP of the gas. Then

the work done in this case will be:

Ext2 1w P (V V ) ...(23)

In adiabatic expansion q=0, therefore from first law equation i.e. U = q + w

Thus w U

Further, V 2 1w U C (T T ) ...(24)

From Equations (23) and (24),

Ext1 2 V 2 1P (V V ) C (T T )

or Ext ExtV 1 2 1 2 1 1 2 2C (T T ) P (V V ) P (RT / P RT / P )

= Ext1 2 2 1 1 2RP (T P T P ) / P P …(25)

Thus, T2 can be calculated from the values of V 1 1 2C , T , P , P and ExtP .

Since the external pressure is constant, work of expansion is given by:

−𝑤𝑖𝑟𝑟 = ∫ 𝑃𝐸𝑥𝑡𝑑𝑉 = 𝑃𝐸𝑥𝑡 (𝑉2 − 𝑉1) = 𝑃𝐸𝑥𝑡 (𝑛𝑅𝑇2

𝑃2−

𝑛𝑅𝑇1

𝑃1)

𝑉2

𝑉1

5. Reversible Isothermal expansion of a Vander Waal’s gas

Now finding the expressions for w, U , H and q for reversible isothermal expansion of a

real gas

5.1 Work of expansion, w.

It is already stated that in the expansion of the gas the work done is given by:

– dw = PdV.

Which can be written as

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V2

V1

w PdV …(26)

Since, the van der Waals gas equation is

2

2

anP (V nb) nRT

V

…(27)

Rearranging the above equation to get the expression for pressure, we get:

2

2

nRT anP

V nb V

…(28)

Substituting equation ( 28) in (26), we get

2 2V V V2 2 2

2 2V V V1 1 1

nRT an nRT anw dV dV dV

V nb V nbV V

22

1 2 1

V nb 1 1w nRT ln an

V nb V V

…(29)

5.2 Internal Energy Change, U

Energy is the function of temperature and volume and it can be represented as:

E = E (T, V)

Taking the differential form:

𝑑𝐸 = (𝜕𝐸

𝜕𝑇)

𝑉𝑑𝑇 + (

𝜕𝐸

𝜕𝑉)

𝑇𝑑𝑉

In the form of heat capacity, the above equation can be written as:

= 𝐶𝑉𝑑𝑇 + (𝜕𝐸

𝜕𝑉)

𝑇𝑑𝑉

For isothermal change, dT = 0

Therefore, 𝑑𝐸 = (𝜕𝐸

𝜕𝑉)

𝑇𝑑𝑉

(𝜕𝐸

𝜕𝑉)

𝑇 is called internal pressure for Vander Waals gas and is given by:

(𝜕𝐸

𝜕𝑉)

𝑇=

𝑎𝑛2

𝑉2

The internal pressure is also given by T( U / V) . Therefore,

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2 2T( U / V) an / V

Rearranging the above equation, change in internal energy at constant temperature will be

2 2dU an (dV / V )

Integrating the above equation, the change in internal energy will be:

V2 2 22 1 2V 2 11

dV 1 1dU U U U an an

V VV

…(30)

5.3 Enthalpy Change, H

The enthalpy is given by , H = U + PV

Let at initial state the enthalpy of the gas is

1 1 1 1H U P V …(31)

The enthalpy at final state is

2 2 2 2H U P V …(32)

So the change in enthalpy is given by:

2 1 2 2 2 1 1 1H H H (U P V ) (U P V )

2 1 2 2 1 1 2 2 1 1(U U ) (P V P V ) U (P V P V ) …(33)

Now considering the equation for van der Waals gas i.e.

2

2

nRT anP

V nb V

Multiplying both sides of the above equation by V:

2nRTV anPV

V nb V

Now substituting this expression of PV in the form of 2 2P V and 1 1P V in equation (33), we get

22 1

2 1 1 2

V V 1 1H U nRT an

V nb V nb V V

…(34)

Solving for 2 1

2 1

V V

V nb V nb

we obtain

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2 1 1 2

2 1

V (V nb) V (V nb)

(V nb) (V nb)

1 2

2 1 2 1

nb (V V ) nb nb

(V nb) (V nb) V nb V nb

…(35)

Substituting equation (35) in equation (34)

2

2 1 1 2

nb nb 1 1H U nRT an

V nb V nb V V

Now substituting the expression for U from equation (30), we get

2 2

2 1 2 1

1 1 1 1H n bRT 2an

V nb V nb V V

…(36)

5.4 Heat Change, q

From the first law of thermodynamics we can find the expression for q i.e.

q = U – w

Since, ∆𝑈 = −𝑎𝑛2 (1

𝑉2−

1

𝑉1)

Now substituting the expressions for w and U obtained from equation (29) and (30) we finally

get

−𝑤 = 𝑛𝑅𝑇 𝑙𝑛 (𝑉2−𝑛𝑏

𝑉1−𝑛𝑏) + 𝑎𝑛2 (

1

𝑉2−

1

𝑉1)

2

1

V nbq nRT ln

V nb

6. Comparison of work of expansion of an Ideal gas and non-ideal gas

The work done in expansion of ideal gas is given by:

- ideal 2 1w nRT ln (V / V )

While that for van der Waals gas the work done is

- 22vdw

1 2 1

V nb 1 1w nRT ln an

V nb V V

…(37)

If V>> nb , the equation (37) becomes

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- 22vdw

1 2 1

V 1 1w nRT ln an

V V V

…(37.1)

Now subtracting equation (37.1) from equation (37) we get

-2

2 2 1ideal vdw

2 1 1 2

an (V V )1 1w ( w ) an

V V V V

…(38)

Since in expansion, final volume V2 is greater than the initial volume V1 therefore using

this in equation (38) we conclude that the reversible work done in the case of ideal gas is greater

than that for van der Waal gas. This can also be explained physically. An ideal gas has no

intermolecular forces whereas a real (van der Waals) gas has considerable intermolecular

forces. Thus, in an ideal gas, the heat supplied is fully utilized in doing the work of expansion

whereas in a real gas, a part of the heat supplied is used in overcoming the intermolecular forces

of attraction and the balance amount of heat is utilized in doing the work of expansion.

Consequently, the work done in the expansion of an ideal gas is numerically greater than the

work done in the expansion of a real gas.

Exercise

Question 1: One mole of benzene is converted reversibly into vapour at its boiling point 80.2ºC

by supplying heat. The vapour expands against the pressure of 1 atm. The heat of vaporization

of benzene is 395 J g-1. Calculate q, w, ΔE and ΔH for the process.

Solution: The process is

𝐶6𝐻6(𝑙𝑖𝑞) 𝐶6𝐻6(𝑣𝑎𝑝𝑜𝑢𝑟)

Heat supplied for conversion of 1 mole of benzene into its vapour.

q = Heat of Vapourization per g * molecular weight of benzene

q = (395 J g-1)*(78 g mol-1)

q = 30810 J mol-1

Work done will be:

-w = PΔV

= P ( Vf – Vi) = P Vf = RT (Vf >> Vi)

T= 273 + 80.2 = 353.2 K

-w = 8.314 * 353.2 = 2936.5 J mol-1

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ΔE = q + w (First law)

= 30810 – 2936.5

= 27873.5 J mol-1

ΔH = ΔE + PΔV = qp = 30810 J mol-1

Question 2: One mole of gas is allowed to expand isothermally and reversibly from a volume

of 1 dm3 to 50 dm3 at 273 K. Calculate w, q, ΔE assuming ideal gas behavior and Vander

Walls behavior (given a = 6.5 atm dm6 mol-2, b = 0.056 dm3 mol-1 and R=0.082 atm dm-3 K-1

mol-1

Solution

(i) Ideal gas

−𝑤 = 2.303 ∗ 𝑅𝑇 ∗ 𝑙𝑜𝑔𝑉2

𝑉1

−𝑤 = 2.303 ∗ (0.082 ∗ 273) 𝑙𝑜𝑔 (50

1)

−𝑤 = 87.6 𝑎𝑡𝑚 𝑑𝑚3 𝑚𝑜𝑙−1

−𝑤 = 8876 𝐽

For isothermal expansion of an ideal gas

ΔE = 0

Therefore, ΔE = q + w

Will give, -w = q = 8876 J mol-1

(ii) Vander Waals gas

−𝑤 = 2.303 ∗ 𝑅𝑇 ∗ 𝑙𝑜𝑔 (𝑉2−𝑛𝑏

𝑉1−𝑛𝑏) + 𝑎𝑛2 (

1

𝑉2−

1

𝑉1)

−𝑤 = 8352 𝐽𝑚𝑜𝑙−1

∆𝐸 = −𝑎 (1

𝑉2−

1

𝑉1)

= −(6.5) (1

50−

1

1) = 645.4 𝐽𝑚𝑜𝑙−1

𝑞 = ∆𝐸 − 𝑤

= 8997.6 𝐽𝑚𝑜𝑙−1

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7. Summary

Work done by an ideal gas in the case of isothermal expansion is

2 1 1 2w nRT ln (V / V ) nRT ln (P / P )

While in the case of isothermal compression is

1 2 2 1w nRTln (V / V ) nRT ln (P / P )

The relation between temperature and respective volume in reversible adiabatic

expansion is given by

11 2 2 1T / T (V / V )

Comparing the change in thermodynamic quantities in the case of ideal gas and real gas,

we get

Work done

V2 2

V 11

VdVw RT RT ln

V V - ideal gas

22

1 2 1

V nb 1 1w nRT ln an

V nb V V

- real gas

Heat change, q

q = -w

Thus, V2 2

V 11

VdVq RT RT ln

V V - ideal gas

2

1

V nbq nRT ln

V nb

- real gas

Enthalpy change , ΔH

ΔH=0 - ideal gas

2 2

2 1 2 1

1 1 1 1H n bRT 2an

V nb V nb V V

- real gas

subject chemistry paper no and title 10, physical chemistry...

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