Solutions to Problem Set 6

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Hugh Chapman
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1 Solutions to Problem Set 6 1. non- ideal gas, 1 mol 20.0 L 300 K 40.0 L 300 K isothermal, reversible Equation of state: (a)b is a constant independent of T Given du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 dt = 0 Given, isothermal 2 du = ( U/ V) T dv 3 Derived from first and second law of thermodynamics 4 ( p/ T) V = {RT[V -1 + BV -2 ]} = R[V -1 + BV -2 ] = T R[V -1 + BV -2 ] - RT[V -1 + BV -2 ] = 0 and using cross derivatives (see lecture notes Part 4) Apply to this equation of state, where B (has units of volume) is indep of T ΔU = 0 From Eq 1, 2 and 5 6 dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) We could do this: ( H/ p) T = - T( V/ T) p + V Derived from first and second law of thermodynamics and using cross derivatives (see lecture notes Part 4) But equation of state is too complex to derive( V/ T) p. Use another route to get ΔH H = U + pv Definition 7 ΔH = ΔU + p f V f -p i V i p = RT/V + BRT/V 2 p i = ( )(300)[(1/20.0) + (B/ )] p f = ( )(300)[(1/40.0) + (B/ )] Δ(pV) = ( )(300)B(-0.025) ΔH = 0+( )(300)B (-0.025) = ( J mol -1 K -1 )(300)B (-0.025) Substituting known values ΔU = q + W First law of thermodynamics 10 q REV = - W REV From Eq 6 and

2 W = - p op dv Definition 12 p op = p gas reversible 13 W REV = - [RT/V + BRT/V 2 ]dv = - RT ln(v f /V i ) +2BRT[(1/V f )-(1/V i )] Using equation of state for this gas and integrating 14 W REV = -RT[ ln(2) + 2B (-0.025)] Substituting 15 = - ( J mol -1 K -1 )(300) [ln(2) B] = - q REV ds = δq REV /T ΔS = q REV /T Second Law of thermodynamics reversible isothermal ΔS = + ( J mol -1 K -1 ) From Eq 15 and [ln(2) B] A = U - TS Definition 19 ΔA = ΔU - TΔS For constant T (this problem) 20 ΔA = ( J mol -1 K -1 ) From Eq 6, 18 and [ln(2) B] G = H - TS Definition 21 ΔG = ΔH - TΔS For constant T (this problem) 22 ΔG = 300 ( J mol -1 K -1 ) [-0.025B - ln(2) B] = 300 ( J mol -1 K -1 ) From Eq 9, 18 and [0.025B - ln(2) ] (b) B is dependent on T B = a + bt + ct 2 Given 24 Just as in deriving Eq 5 but with the additional term 25 in (db/dt) Where db/dt = b + 2cT ( p/ T) V = {RT[V -1 + BV -2 ]} = R[V -1 + BV -2 ] + RV -2 (db/dt) = RT[V -1 + BV -2 ] - RT[V -1 + BV -2 ] +RTV -2 (db/dt) ( U/ V) T = RTV -2 (db/dt) ΔU = ( U/ V) T dv = RT(dB/dT) V -2 dv = RT(dB/dT)[(1/V f ) (1/V i )] = ( J mol -1 K -1 )(300) (b + 600c) [ ] p = RT/V + BRT/V 2 p i = ( )(300)[(1/20.0) + (B/ )] p f = ( )(300)[(1/40.0) + (B/ )] Δ(pV) = ( )(300)B(-0.025) From Eq 3 and 25 Note that ΔU 0 when B is dependent on T The equation for Δ(pV) is the same as when B is independent of T 26 27

3 ΔH = ( J mol -1 K -1 )(300) (b + 600c) (-0.025) + ( J mol -1 K -1 )(300)B (-0.025) The same as in part (a) except that ΔU from Eq 26 and B = a + bt + ct 2 are inserted. 28 = - ( J mol -1 K -1 )(300) [a+bt+ct 2 + b + 600c] ΔS = + ( J mol -1 K -1 ) [ln(2) B] ΔA = ΔU - TΔS ΔA = ( J mol -1 K -1 )(300) (b + 600c) ( ) 300 ( J mol -1 K -1 ) [ln(2) B] ΔG = ΔH - TΔS ΔG = - ( J mol -1 K -1 )(300) [a+bt+ct 2 + b + 600c] 300 ( J mol -1 K -1 ) [ln(2) B] ΔG = + ( J mol -1 K -1 )(300) [0.025 (a+bt+ct 2 ) (b + 600c)- ln(2)] (c) B is a constant independent of temperature and the process is carried out irreversibly ΔU = 0 ΔH = - ( J mol -1 K -1 )(300)(0.025)B ΔS = + ( J mol -1 K -1 ) [ln(2) B] ΔA = 300 ( J mol -1 K -1 ) [ln(2) B] ΔG = 300 ( J mol -1 K -1 ) [0.025B - ln(2) ] But q IRREV and W IRREV are not the same as in part (a). q IRREV + W IRREV = 0 q IRREV < TΔS W IRREV > W REV W IRREV > -TΔS The equation is derived the same as in part (a) except that B = a + bt + ct 2 is inserted The same as in part (a) except that ΔU from Eq 26 and B = a + bt + ct 2 are inserted. ΔH from Eq 28 and B = a + bt + ct 2 are inserted. 31 All are exactly the same as in part (a) because these are state functions and do not depend on the path (reversible or otherwise). What do we know about the non-state functions? Because ΔU = 0 here because of Clausius inequality: TΔS > q IRREV Algebraic relation! Remember W is a signed quantity. Negative value means work is done by system on the surroundings. Less work than maximum work appears in the surroundings for irreversible process 29 30

4 2. non- ideal gas, 2 mol 10 atm 600 K 5 atm 300 K irreversible Equation of state: pv = nrt + nbp B = L mol -1 J mol -1 K -1 du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 Derived from first and second law of thermodynamics 2 and using cross derivatives (see lecture notes Part 4) p = nrt/(v-nb) Given equation of state 3 ( p/ T) V = nr/(v-nb) Apply to this equation of state, where B (has units of volume) is independent of T 4 = nrt/(v-nb) - nrt/(v-nb) = 0 du = C V dt + 0 From Eq 1 and 4 5 General relation derived from U = U(T,V) and the first 6 law of thermodynamics (see lecture notes Part 2) V-nB = nrt/p Rearrange equation of state and differentiate 7 ( V/ T) p = nr/p C p -C V = [p +0](nR/p) = nr From Eq 6, 7 and 4 8 C V = C p nr = n[ From Eq 8 and given C p T ] J mol -1 K -1 C V = 2 mol ( T) J mol -1 K -1 ΔU = 2 mol Integrating Eq 5 and using given C V 10 ( T)dT ΔU = 2{44.361(T 2 -T 1 ) + (1/2) (T 2 2 -T 2 1 )} = 2{(44.361( ) + (1/2) ( )} = 2{ } Integrating Eq 5 and using given C V 11 = J dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 12 ( H/ p) T = - T( V/ T) p + V. Derived from first and second law of thermodynamics 13 and using cross derivatives (see lecture notes Part 4) ( V/ T) p = nr/p and V = nrt/p +nb From Eq 7 and Equation of state substituted into Eq 13 ( H/ p) T = nb 14 dh = C p dt + nb dp ΔH = n( T)dT + nbdp From Eq 12 and 14 15

5 ΔH = 2{52.675(T 2 -T 1 ) + (1/2) (T 2 2 -T 2 1 )} L mol -1 [p 2 -p 1 ] = 2{(52.675( ) + (1/2) ( )} [5-10] / = 2{ } = J Integrating and substituting known values We would have obtained the same answer by using H = U+pV ΔH = ΔU + Δ(pV) It happens that dv=0 here ds = ( S/ T) p dt + ( S/ p) T dp S = S(T,p) 12 Derived (see lecture notes Part 4) starting from 13 and use of cross derivatives ds = (C p /T)dT - nrdp/p Using Eq 7 and 2 into Eq ΔS = n( /T )dt Using Eq 7 and 2 into Eq 12 - nr dp/p 15 ΔS = 2{ Integrating and substituting given values ln(T 2 /T 1 ) (T 2 -T 1 ) ln(p 2 /p 1) } ΔS = 2{ ln(300/600) ( ) We would have obtained the same answer by using ln(5/10) } ds = ( S/ T) V dt + ( S/ V) T dv ( S/ V) T = ( p/ T) V but it happens that dv=0 here, so ΔS = J K -1 ΔS = C V dt/t = same answer A = U TS ΔA = ΔU - Δ(TS) S abs, f = S abs,i + ΔS Δ(TS) = 300S abs,i - 600[S abs, i + ΔS ] ΔA = J - 300S abs, i - 600[S abs, i ] G = H -TS ΔG = ΔH - Δ(TS) S abs, f = S abs,i + ΔS Δ(TS) = 300S abs,i - 600[S abs, i + ΔS ] ΔG = J - 300S abs, i - 600[S abs, i ] Definition We already have ΔU. Since dt is not zero for this problem, it requires that we know absolute entropy. We have to have either the absolute entropy for the gas at the initial state or at the final state in order to calculate ΔA. If we are given data to calculate S abs, i, then we can do it. Definition We already have ΔH. As from above argument, we have to use the absolute entropy for the gas at the initial state and at the final state in order to calculate ΔG 8

6 3. non- ideal gas, 2 mol p i =1 atm T i =300 K p f =11 atm T f =300 K V i = a + RT i /p i V f = a + RT f /p f Equation of state: pv = RT +ap a = a(t) du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 Derived from first and second law of thermodynamics 2 and using cross derivatives (see lecture notes Part 4) pv = RT +ap Given equation of state 3 p = RT/(V-a), a=a(t) ( p/ T) V = R/(V-a) + RT(da/dT)/(V-a) 2 Apply to this equation of state 4 = p + RT 2 (da/dt)/(v-a) 2 - p ( U/ V) T = RT 2 (da/dt)/(v-a) 2 Q.E.D. Using Eq 2 5 dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 6 ( H/ p) T = - T( V/ T) p + V. Derived from first and second law of thermodynamics 7 and using cross derivatives (see lecture notes Part 4) V = RT/p + a From Equation of state (Eq 3) ( V/ T) p = R/p + da/dt 8 ( H/ p) T = - T( V/ T) p + V = -RT/p -T(da/dT) + RT/p + a Substituting Eq 8 into Eq 7 ( H/ p) T = a -T(da/dT) Q.E.D. 9 We will need da/dt in Eq 5 and Eq 9 10 a = (T-300) L mol -1 The given 3 values of a(t) changes linearly with T by per 100 K increment (da/dt) = L mol -1 K -1 du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 du = C V dt + RT 2 (da/dt)/(v-a) 2 dv Substituting Eq 5 into Eq 1 11 In this problem dt = 0 ΔU = 2 mol Integrating Eq 11 RT 2 (da/dt) (V-a) - 2 dv = 2 mol RT 2 (da/dt) (-1)(V-a) -1 f i 12 = - 2 mol RT 2 f (da/dt) p/rt i Using equation of state to express (V-a) -1 = - 2 mol T(da/dT) [p f - p i ] 13 ΔU = - 2 mol (300 K)( L substituting values mol -1 K -1 ) [11-1] atm Using conversion factor J / L atm ΔU = L atm or J 14 dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 6 dh = 0 + {a - T(da/dT)}dp Using dt = 0 and Eq 9 into Eq 6 15

7 ΔH = 2 mol {0.02 L mol -1 - (300 K)( L mol -1 K -1 )} [p f - p i ] ΔH = 2 mol L mol -1 [11-1] atm ΔH = L atm or J W = - p op dv W max = - p gas dv = -2 mol RT (V-a) -1 dv = -2 mol RT ln [(V f - a)/(v i - a)] = -2 mol RT ln (p i /(p f ) W max = - 2 mol ( J mol -1 K -1 )(300K)ln(1/11) W max = J dg = Vdp - SdT dt = 0 for this process ΔG = Vdp = 2 mol (RT/p + a)dp ΔG =2 mol { RTln(p f /pi) + a[p f -pi] } = 2 mol {( J mol -1 K -1 )(300K) ln(11/1) L mol -1 [11-1] atm J / L atm} ΔG = J J ΔG = J da = - pdv - SdT dt = 0 for this process ΔA = - pdv = - RTdV/(V-a) = - 2 mol RT ln [(V f - a)/(v i - a)] = - 2 mol RT ln (p i /(p f ) ΔA = - 2 mol ( J mol -1 K -1 )(300K)ln(1/11) ΔA = J Integrating Eq 15 Substituting values Maximum work is obtained by carrying out the process reversibly, in which case p op = p gas Integrating Using the equation of state Substituting values one of the fundamental equations of thermodynamics (Lecture notes Part 5) Using the equation of state Integrating one of the fundamental equations of thermodynamics (Lecture notes Part 5) Integrating Using the equation of state Substituting the values Note that ΔA = W max, as it should be

4. 0.5 mol CH 4 (g) at 298 K is burned in excess O 2 under adiabatic conditions. T final =?

8 mol CH 4 (g) at 298 K is burned in excess O 2 under adiabatic conditions. T final =? The reaction at 298 K CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Standard enthalpy of combustion at 298 K = Δ comb H 298 = 2Δ form H 298 (H 2 O(l)) +Δ form H 298 (CO 2 (g)) - Δ form H 298 (CH 4 (g)) - 2Δ form H 298 (O 2 (g)) Δ comb H 298 = 2Δ form H 298 (H 2 O(l) ) +Δ form H 298 (CO 2 (g) ) - Δ form H 298 (CH 4 (g) ) = 2( )+( )-(-74.81) kj mol -1 Δ comb H 298 = kj mol -1 Δ form H 298 for the elements in their standard state is defined as zero Using Δ form H 298 given in the provided table for CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) all at 298 K 1 2 (a) at constant pressure ~ 1 bar q p = 0 adiabatic (insulated container and piston) 0.5 mol CH mol CO 2 (g) 10 mol O 2 [10-1] mol O 2 (g) 0.5(2) mol H 2 O(g) 298 K T f =? dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 3 dh = C p dt + 0 ΔH = q p q p = 0 Given dp = 0 Given adiabatic ΔH = ½ Δ comb H Tf 298 C p (CO 2 g)dt + 9 Tf 298 C p (O 2 g)dt C p (H 2 Oliq)dT + 1 Δ vap H 373 (H 2 O) + 1 Tf 373 C p (H 2 Og)dT = 0 Need to sum up all the contributions to ΔH 7

0.5(- 890.36 10 3 ) + 0.5(37.11)(T f - 298) + 9(29.355)(T f - 298) +1(75.291)(373-298) + 1(40.656 10 3 ) + 1(33.

9 0.5( ) + 0.5(37.11)(T f - 298) + 9(29.355)(T f - 298) +1(75.291)( ) + 1( ) + 1(33.58)(T f - 373) = 0 Substitute values of ΔH in J mol -1 from Eq 2 and into Eq 7 and look up C p in J mol -1 K -1 for CO 2 (g), O 2 (g), H 2 O(liq) and H 2 O(g) in given table T f = 1567 K Solving Eq 8 for T f 9 (b) at constant volume adiabatic (insulated container) with rigid walls 8 initial final 0.5 mol CH mol CO2(g) 10 mol O 2 [10-0.5] mol O2(g) 0.5(2) mol H 2 O(g) 298 K T f =? du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 10 du = ( U/ T) V dt + 0 Given dv = 0 11 ΔU = q V 12 q V = 0 Given adiabatic 13 ΔU = ½ Δ comb U 298 Need to sum up all the contributions to ΔU Tf 298 C V (CO 2 g)dt + 9 Tf 298 C V (O 2 g)dt C V (H 2 Oliq)dT + 1 Δ vap U 373 (H 2 O) + 1 Tf 373 C V (H 2 Og)dT = 0 H = U + pv Definition 15 ΔH = ΔU + Δ(pV) 16 Δ comb H 298 = Δ comb U 298 Apply Eq 16 to the reaction CH 4 (g) + 2O 2 (g) 17 +(pv) products - (pv) reactants (pv) products = (298 R) 1 (pv) reactants = (298 R) 3 (pv) products - (pv) reactants = - 2(298 R) = - 2(298)( ) (pv) products - (pv) reactants = J mol -1 Δ comb U 298 = (-4955) = J mol -1 CO 2 (g) + 2H 2 O(l) at 298 K Assuming ideal gas behavior for reactants and products, neglecting pv for liquid 19 Substituting the values from Eq 2 & 19 into Eq

10 Δ vap H 373 = Δ vap U (pv) gas - (pv) liq Δ vap U 373 = Δ vap H Δ(pV) Δ(pV) = (373R) = 3101 J mol -1 Δ vap U 373 = ( ) Substituting the values = J mol -1 ΔU = ½( ) + 0.5(28.8)(T f - 298) + 9(21.0)(T f - 298) +1(75.2)( ) + 1( ) + 1(25.3)(T f - 373) = 0 T f = 2053 K Solving Eq 26 for T f Note that this is a higher final temperature than the final temperature reached in doing the combustion at constant pressure. Apply Eq 16 to vaporization 22 Assuming ideal gas behavior for gas and neglecting pv for liquid Using values for Eq 21 & 25 into Eq 14 and looking up C V values in J mol -1 K -1 for CO 2 (g), O 2 (g), H 2 O(liq) and H 2 O(g) in given table This is because all of the chemical energy released by the reaction at constant volume went into raising the temperature of the products, whereas, some of the chemical energy released by the reaction at constant pressure went into Work against p op = 1 bar The formation of H 2 O(g) at 298 K H 2 (g) + ½O 2 (g) H 2 O(g) given Δ form H 298 = kj mol -1 Initial Final 1 mol H 2 (g) (1-x) mol H 2 (g) 10 mol O 2 (g) (10-½x) mol O 2 (g) x mol H 2 O(g) T = 298 T = 2000 K constant pressure set up the problem as shown above Recognize that the reaction is that for the formation of H 2 O(g) dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 1 dh = C p dt + 0 ΔH = q p Given dp = q p = 0 Assume that all the Δ rxn H is accounted for by the final 4 temperature of 2000 K, i.e., as if H 2 was burned in an insulated cylinder/piston at ~ 1 atm or 1 bar. ΔH = x Δ form H (1-x) C p (H 2 g)dt + (10-½x) C p (O 2 g)dt + x C p (H 2 Og)dT = 0 Need to sum up all the contributions to ΔH 5

11 x ( ) + (1-x) (28.824)( ) + (10-½x)(29.355)( ) + x (33.58)( ) = 0 x = 0.47 fraction of original H 2 that had undergone combustion Substitute value of Δ form H 298 in J mol -1 into Eq 5 and look up C p in J mol -1 K -1 for H 2 (g), O 2 (g), and H 2 O(g) in given table Solving Eq 6 for x f /p of ethane as a function of pressure at 600 K Z = Given at 600 K (p/1 bar) (p/1bar) (p/1bar) (p/1bar) 4 ln(f /p) = p 0 [(Z-1)/p] T dp which we derived in class (see lecture notes Part 5) 2 (Z-1)/p = From Eq (p/1 bar) (p/1bar) (p/1bar) 3 ln(f /p) = p 0 [(Z-1)/p] T dp = (p/1 bar) +½ (p/1bar) 2 - (1/3) (p/1bar) 3 - (1/4) (p/1bar) 4 Substituting Eq 3 into Eq 2 and integrating 4 (f /p) = exp{ (p/1 bar) +½ (p/1bar) 2 - (1/3) (p/1bar) 3 - (1/4) (p/1bar) 4 } at 600 K Rearranging Eq 4 5

12 7. Distribution of a real gas in a gravity field F = mg F z+dz + df = F z See lecture notes Part 1 for starting this problem Pressure at any height z is determined by the column 1 2 of fluid above that height. Start withforce arising from the weight of fluid in the column at z (F z ) and compare that with the force arising from the weight of the fluid in the column at z+dz. Let ρ be the density at height z The mass of fluid in the column between z and z+dz is ρadz A is the cross sectional area of the column df = (ρadz)g 3 The pressure: p at height z = F z /A p + dp at height z+dz = F z+dz /A A(p+dp) +df = Ap df = -Adp Substituting Eq 4&5 into Eq 2 Rearranging 6 dp = - ρgdz Substituting Eq 6 into Eq 3 7 ρ = M/V where M is the molar mass and V is the molar volume dp = - (M/V) gdz substituting for ρ in Eq 7 8 dp/p = - (M/ZRT) gdz Using Z = pv/rt or V = ZRT/p into Eq 8 9 p p0 (dp/p) = - (M/ZRT) g z 0 dz ln(p/p 0 ) = - (Mg/ZRT) z p = p 0 exp[- (Mg/ZRT) z] (a) if Z > 1, exp[- (Mg/ZRT)z] > exp[- (Mg/RT)z] That is, need to go to larger height to drop from p 0 (at the surface) to the same pressure p it would have been had the gas been ideal. This means the gas distribution is broader when Z > 1. (b) if Z < 1, exp[- (Mg/ZRT)z] < exp[- (Mg/RT)z] That is, don t need to go as high to drop from p 0 to the same pressure p it would have been had the gas been ideal. This means the gas distribution is narrower Z < 1. Integrating between z = 0 and z 10 Comparing with ideal case where Z = 1 If Z = 1+Bp dp/p = - [M/(1+Bp)RT] gdz Substitute this into Eq 9 {(1+Bp)/p} dp = - [Mg/RT] dz Rearranging

13 p0 p (dp/p) + B p0 p dp = - (Mg/RT) 0 z dz ln(p/p 0 ) + B(p-p 0 ) = - (Mg/RT) z Integrating between z = 0 and z to be compared with the ideal case: ln(p/p 0 ) = - (Mg/RT) z 8. Fugacity of a gas Z = 1 + [b-(a/rt)](p/rt) (Z-1)/p = [b-(a/rt)]/rt Given Rearranging 1 2 ln(f /p) = p 0 [(Z-1)/p] T dp which we derived in class (see lecture notes Part 5) 3 ln(f /p) = p 0 {b-(a/rt)]/rt} T dp = [b-(a/rt)]/rt} T p Substituting Eq 2 into Eq 3 Integrating (f /p) = exp[(p/rt)[b-(a/rt)] f = p exp[(p/rt)[b-(a/rt)] 9. The actual chemical potential is Recall that the standard state for a gas corresponds 1 μ(t,p) = μ (T) + RTln (f/1bar) to a fictitious state of an ideal gas at 1 bar and temperature T for which the chemical potential is μ (T) μ(t,p) = μ (T) + Rln (f/1bar) Dividing Eq 1 by T 2 T T (1/n) G = (1/n) G + Rln (f/1bar) Recall that for a substance μ = G/n where n is the 3 T T number of moles (1/n)[ / T(G/T)] p = (1/n)[ / T(G /T)] p + R ln (f/1bar)/ T] p Take the derivative of both sides with respect to T at constant p [ / T(G/T)] p = - H/T 2 The Gibbs-Helmholtz equation (see lecture notes Part 5) - H/T 2 = - H /T 2 +R ln (f/1bar)/ T] p Substitute Eq 5 into Eq 4, where H and H are now molar quantities H = H - R T 2 ln (f/1bar)/ T] p H is H ideal since the standard state at temperature T Q.E.D. corresponds to a fictitious gas behaving ideally at 1 bar, whereas H is the molar enthalpy of the real gas

The Second Law of Thermodynamics (Chapter 4)

The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made