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
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
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
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
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.
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
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 wP dV …(4) V1 Substituting P = RT/V in the above expression for one mole of an ideal gas thus giving,
V2 dV V wRTRT ln 2 …(5) V1 VV1
At constant temperature, P1122 VP V for the case of ideal gas, thus work done can also be written as
w RT ln (P12 / P ) …(6) For n moles of the gas, the work done can be written as:
푃1 w nRTln(V2 /V) 1 nRTln(P/P) 1 2 =−2.303 푛푅푇 푙표푔 …(7) 푃2 CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
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.
푛푖 푐1 Since 푐푖 = therefore, 푤 = −푛푅푇 푙푛 푉 푐2
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(PdP) dVPdV …(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 wP dV V2
Taking the gas to be ideal, thus substituting P= RT/ V and PVPV1122 in the above expression we get,
V1 dV w RT RTln(V/V)12= RT ln (P21 / P ) …(9) V2 V For n moles of the gas, the work done is given by:
w nRTln(V/V)1 2 nRTln(P 2 /P) 1 …(10)
Since during compression the initial volume V2 > V1 and P2 < P1 thus w’ will always be positive.
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
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 wdwPdV0ext …(11) Intermediate expansion ext Let the volume expands from V1 to V2 against an external pressure P then the work done will be V 2 extext wPdVP(VV ) 21 …(12) V1 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), 푛푅푇 푛푅푇 Therefore, −푤 = 푃푒푥푡 ( − ) 푃2 푃1 푃2 ext −푤 = 푛푅푇 (1 − ) (if P2 ≈ P ) 푃1
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
U0w or wU …(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 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.
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
Calculation of U . At constant volume the molar heat capacity for an ideal gas is given by:
C(U/T)VV …(14) Accordingly finite change in internal energy will be
U C d T V …(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
HU(PV)URT …(16)
Substituting the value of , we get
HCTRVV T(CR)T = CTP …(17) Calculation of w In adiabatic process q=0, thus from the first law equation = q + w, the work done will be given by:
wUCT V Thus from the above equations we can say that all the three quantities , 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 . Hence, according to the First law equation,
= – P …(18) CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
Let the fall in temperature be T and U = CTV
Therefore, CTV = – PdV ...(19)
For infinitesimally small quantities,
CdTPdVRTV dV / V (for 1 mole of the gas) Dividing the above equation by T
CdT/TRdV/VV
Or Cd(lnT)Rd(lnV V)
퐶 Or 푣 푑(ln 푇) = −푑(ln 푉) ...(20) 푅
Integrating the above equation between the limits of temperatures T1 and T2 with the
corresponding volumes V1 and V2 then we get,
Cln(TV212112 /T)Rln(V /V)Rln(V /V ) ...(21)
or ln (T/21V12 T )(R / C) ln (V / V )
Since, CCRPVand substituting C/CPV, we get
푇 퐶 푉 푙푛 ( 2) = 푣 = 푙푛 ( 1) (for one mole) 푇1 푅 푉2
푛푅 푅 푉1 퐶푣 푉1 퐶푣,푚 푇2 = 푇1 ( ) = 푇1 ( ) 푉2 푉2 훾−1 푉1 푇2 = 푇1 ( ) …(22) 푉2
1 or T/1221 T(V/ V ) (Note that > 1)
Since in the case of expansion V2 V 1 and T 2 T 1 . 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 CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
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 , as and are zero therefore is also zero. Thus, in a free adiabatic expansion, = 0, w = 0 and = 0. Intermediate Expansion.
E xt Let the gas expands from the volume V1 to V2 against a constant pressure P of the gas. Then the work done in this case will be:
Ext wP(VV) 21 ...(23) In adiabatic expansion q=0, therefore from first law equation i.e. = q + w
Thus wU
Further, wUC(TT) V21 ...(24) From Equations (23) and (24),
Ext P(VV)C(TT12V21 )
ExtExt or C(TTV12121122 )P(VV )P(RT / PRT / P )
Ext = RP(T PT122 P ) 112 / P P …(25)
E xt Thus, T2 can be calculated from the values of C,T,P,PV112 and P . Since the external pressure is constant, work of expansion is given by:
푉2 푛푅푇 푛푅푇 −푤 = ∫ 푃퐸푥푡푑푉 = 푃퐸푥푡 (푉 − 푉 ) = 푃퐸푥푡 ( 2 − 1) 푖푟푟 2 1 푃 푃 푉1 2 1
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
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
V2 w PdV …(26) V1 Since, the van der Waals gas equation is
an2 P(Vnb)nRT …(27) 2 V Rearranging the above equation to get the expression for pressure, we get:
nRTan 2 P …(28) Vnb V2 Substituting equation ( 28) in (26), we get
VVV222nRTannRTan22 wdVdVdV 22 VVV111VnbVnbVV
Vnb2 2 11 wnRT lnan …(29) VnbVV121
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:
휕퐸 휕퐸 푑퐸 = ( ) 푑푇 + ( ) 푑푉 휕푇 푉 휕푉 푇 In the form of heat capacity, the above equation can be written as:
휕퐸 = 퐶푉푑푇 + ( ) 푑푉 휕푉 푇 For isothermal change, dT = 0
휕퐸 Therefore, 푑퐸 = ( ) 푑푉 휕푉 푇
휕퐸 ( ) is called internal pressure for Vander Waals gas and is given by: 휕푉 푇
휕퐸 푎푛2 ( ) = 2 휕푉 푇 푉
The internal pressure is also given by (U /V) T . Therefore, CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
22 (U/V)an/V T Rearranging the above equation, change in internal energy at constant temperature will be dUan(dV/V) 22
Integrating the above equation, the change in internal energy will be:
V2 dV11 dUUUUanan 22 …(30) 21 2 V1 V VV21
5.3 Enthalpy Change, H The enthalpy is given by , H = U + PV Let at initial state the enthalpy of the gas is
HUPV1111 …(31) The enthalpy at final state is
HUPV2222 …(32) So the change in enthalpy is given by:
HHH(UP2122 V 211)(UP 1 V )
(UU212 )(P 21 VP 12 V 21 )U(P 1 VP V ) …(33) Now considering the equation for van der Waals gas i.e.
nRTan 2 P Vnb V2 Multiplying both sides of the above equation by V:
nRTVan 2 PV Vnb V
Now substituting this expression of PV in the form of PV22 and PV11 in equation (33), we get
VV212 11 HU nRTan …(34) Vnb2112 V nbVV
VV Solving for 21 we obtain VnbVnb21
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
V(Vnb)V(Vnb) 2112 (Vnb)21 (Vnb)
nb (VV) nbnb 12 …(35) (Vnb)2121 (Vnb)VnbVnb Substituting equation (35) in equation (34)
nbnb11 2 HUnRTan VnbVnbVV2112
Now substituting the expression for U from equation (30), we get
22 1111 HnbRT2an …(36) VnbVnbVV2121 5.4 Heat Change, q From the first law of thermodynamics we can find the expression for q i.e.
q = U – w
1 1 Since, ∆푈 = −푎푛2 ( − ) 푉2 푉1
Now substituting the expressions for w and U obtained from equation (29) and (30) we finally get
푉 −푛푏 1 1 −푤 = 푛푅푇 푙푛 ( 2 ) + 푎푛2 ( − ) 푉1−푛푏 푉2 푉1
Vnb2 qnRT ln Vnb1
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:
- wnRTideal21 ln (V / V ) While that for van der Waals gas the work done is
V2 nb 2 11 - wvdw nRT ln an …(37) V1 nb V 2 V 1 If V>> nb , the equation (37) becomes
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
V2 2 11 - wvdw nRT ln an …(37.1) VVV1 2 1 Now subtracting equation (37.1) from equation (37) we get
2 2 11an(VV21 ) - w(w)anidealvdw …(38) VVV2112 V
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
퐶6퐻6(푙푖푞) 퐶6퐻6(푣푎푝표푢푟) 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
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
ΔE = q + w (First law) = 30810 – 2936.5 = 27873.5 J mol-1
-1 ΔH = ΔE + PΔV = qp = 30810 J mol 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
푉 −푤 = 2.303 ∗ 푅푇 ∗ 푙표푔 2 푉1
50 −푤 = 2.303 ∗ (0.082 ∗ 273) 푙표푔 ( ) 1 −푤 = 87.6 푎푡푚 푑푚3 푚표푙−1 −푤 = 8876 퐽 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
푉 −푛푏 1 1 −푤 = 2.303 ∗ 푅푇 ∗ 푙표푔 ( 2 ) + 푎푛2 ( − ) 푉1−푛푏 푉2 푉1 −푤 = 8352 퐽푚표푙−1
1 1 ∆퐸 = −푎 ( − ) 푉2 푉1
1 1 = −(6.5) ( − ) = 645.4 퐽푚표푙−1 50 1 푞 = ∆퐸 − 푤 = 8997.6 퐽푚표푙−1
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion
7. Summary
Work done by an ideal gas in the case of isothermal expansion is
w nRTln(V2 /V) 1 nRTln(P/P) 1 2 While in the case of isothermal compression is
wnRTln (V/ V)nRT1221 ln (P/ P ) The relation between temperature and respective volume in reversible adiabatic expansion is given by 1 T/T(V/V)1221 Comparing the change in thermodynamic quantities in the case of ideal gas and real gas, we get Work done
V2 dV V wRTRT ln 2 - ideal gas V1 VV1
Vnb2 2 11 wnRT lnan - real gas VnbVV121 Heat change, q q = -w
V2 dV V Thus, qRTRT ln 2 - ideal gas V1 VV1
Vnb2 qnRT ln - real gas Vnb1 Enthalpy change , ΔH ΔH=0 - ideal gas
22 1111 H n bRT2an - real gas Vnb2121 V nbVV
CHEMISTRY Paper No. 10: Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics) Module No. 4: Isothermal and Adiabatic expansion