CHE 112F Problem Set #7 Solutions November 9, 2016 1. Problem 10–8 2000 J of work is used to compress 1.00 mol of an ideal gas isothermally at 0.00°C from 0.50 atm pressure to 1.00 atm pressure. A 10.0°C temperature drop is provided between the gas and the surroundings to facilitate the heat transfer. Calculate the minimum entropy change in the universe for the compression under these conditions. Solution Taking the gas as the system: ∆Usyst = Q + W = 0 [isothermal, ideal gas] Since, during the compression, Wsyst is positive, then Qsyst is negative (i.e., heat flows out of the system, into the surroundings). Therefore, the temperature of the surroundings is lower than that of the gas, i.e., Tsurr = 263.15 K. "V %
For the compression, W = − ∫ PdV = − nRT ln $ 2 ' # V1 &
= – (1)(8.314)(273.15) ln ( 12 ) = +1574.16 J Q syst = – W syst = –1574.16 J and −1574.16 J
!Q
∆Ssyst = " #$ = = –5.763 J K–1 T syst
273.15 K
For the compression, the actual work done on the system is Wsyst = 2000 J [Note that Wsyst > W syst ] Therefore, Qsyst = –Wsyst = –2000 J and Qsurr = – Qsyst = – (– 2000 J) = + 2000 J The minimum entropy increase for the universe occurs when the process in the surroundings is reversible, i.e., when Q surr = Qsurr = + 2000 J Under these conditions, ∆Ssurr = = +2000 J
= + 7.6002 J K–1 263.15 K
and (∆Suniv)min = ∆Ssyst + ∆Ssurr = – 5.763 + 7.600 = + 1.837 J K–1 Ans: +1.84 J K–1 2. Problem 10–14 100 g of ice at 0.00°C is added to 100 g of water at 100.0°C in an insulated container at a constant pressure of one atm. The surroundings is maintained at a temperature of 300K. When thermal equilibrium has been established, the final temperature of the system is Tƒ kelvins. (a) What is Tƒ? (b) What is ∆Suniv for this process? Data: Specific heat of fusion of ice: ∆ h fus = 335 J K–1 Specific heat capacity of water: c P = 4.18 J g–1 K–1 Solution (a) The 100 g of ice (mass m1) at 0°C will melt to water at 0°C (∆H1); then the 100 g of water at 0°C will heat up to some final system temperature Tƒ (∆H2). Under the cooling influence of the ice, the 100 g of hot water (mass m2) initially at 100°C will be cooled down to Tƒ, the final temperature of the system (∆H3). Since the system is insulated, all enthalpy changes are kept within the system, so that there is no net change in enthalpy within the system; therefore, ∆H1 + ∆H2 + ∆H3 = 0 That is, (m1∆ h fus)1 + (m1 c P ∆T)2 + (m2 c P ∆T)3 = 0 (100)(335) + (100)(4.18)(tƒ – 0) + (100)(4.18)(tƒ – 100) = 0 Solving: tƒ = 9.928°C → Tƒ = 283.078 K Ans: 283 K (b) Since the system is insulated, Qsurr = 0, and ∆Ssurr = 0; therefore, ∆Suniv = ∆Ssyst = ∆S1 + ∆S2 + ∆S3 = m 1Δhfus
" Tƒ $
" Tƒ $
+ m1 cP ln
+ m 2 cP ln
# 273.15 %
# 373.15 %
Tfus
= 33 500
! 283.078 #
! 283.078#
+ (100)(4.18) ln "
+
(100)(4.18)ln
" 373.15 $ 273.15
273.15 $
= +122.64 + 14.92 – 115.48 = + 22.08 J K–1 3. Problem 10–17 Assuming methane behaves as an ideal gas, calculate ∆S for the following process: 1 mol CH 4 #
! 1 mol CH4 # → !
" 0°C, 25.0 kPa$
" 174.8°C, 500 kPa$ At 25 kPa the heat capacity for gaseous methane (with T in kelvins ) is cP = 14.15 + 75.50 × 10–3 T – 17.99 × 10–6 T2 J mol–1 K–1 Solution For one mol of an ideal gas, ∆S = ∫
!P $
cP
dT – R ln # 2 & T
" P1 %
Therefore (we just use the integrated form of the equation again) ! 447.95 #
∆S = (14.15) ln "
+ 75.50 × 10–3 (447.95 – 273.15) – 273.15 $
17.99 × 10 −6
! 500 #
(447.952 − 273.152 ) – (8.314) ln "
25.0 $
2
= (6.9994 + 13.1974 – 1.1338) – 24.9065 = 19.0630 – 24.9065 = – 5.8435 J mol–1 K–1 4. Problem 10–27 Two blocks of copper, each of 1.00 kg mass, one at 1000 K and the other at 200 K, are brought into contact inside an insulated container and allowed to reach thermal equilibrium at atmospheric pressure. What is ∆S for the universe as a result of this process? For copper, c P = 0.385 J g–1 K–1 Solution Let the system consist of the two blocks of copper. Since the container is insulated, Qsyst = 0, and no heat is transferred to the surroundings; therefore ∆Ssurr = 0. The final temperature of each block will be 12 (100 + 200) = 600 K. ! 600 #
! 600 #
∆Ssyst = ΔShot + ΔScold = CP ln "
+ CP ln "
$
1000
200 $
block
block
= CP ( − 0.5108 + 1.0986) = 0.5878CP CP !
J $
(1000 g ) = 385 J K–1 = # 0.385
"
g K%
Therefore, ∆Ssyst = (0.5878)(385) = 226.3 J K–1 and ∆Suniv = ∆Ssyst + ∆Ssurr = + 226.3 + 0 = 226.3 J K–1 Ans: 226 J K–1 5. Problem 10–31 150 g of water at 25°C and 10.0 g of ice at – 15°C are mixed in a thermos bottle. Assuming that the process is adiabatic, determine (a) the final temperature and (b) ∆Suniv. Data: c P [H2O(liq)] = 4.18 J g–1 K–1; c P [ice] = 2.04 J g–1 K–1; Δ hfus [ice] = 333 J g–1 at 0°C Solution (a) Let the final temperature (°C) be Tf . The total change in enthalpy of the ice plus the total change in enthalpy of the water must add to zero, since the system is adiabatic and no heat can leave. Assuming the ice melts (if incorrect this can be checked later), the process for the ice is ΔH2
ΔH1
" 10.0 g water $
" 10.0 g ice
" 10.0 g ice
" 10.0 g water $ ΔH3
'
'
'
'
→
# −15°C $% ''''→ # 0°C $% ''''→ #
%
Tf °C
0°C
#
% ΔH4
! 150 g water #
! 150 g water
The process for the water is " 25°C #$ &&&&→ " T °C $ f
Therefore, ∆Hice + ∆Hwater = 0 . . . [a] !
J $
(0 − (−15)) = 306 J ∆H1 = CP [ice]. ∆Tice = (10.0 g )# 2.04
"
g K%
!
J$
∆H2 = (10.0 g )# 333
= 3330 J "
g%
!
J $
∆H3 = CP [water]. ∆Twater = (10.0 g )# 4.18
(T − 0) = 41.8Tf "
g K% f
!
J $
∆H4 = CP [water]. ∆Twater = (150 g )# 4.18
(T − 25) = 627Tf – 15 675 J "
g K% f
Substituting [b], [c], [d], and [e] into [a]: Solving: . . . [b] . . . [c] . . . [d] . . . [e] 306 + 3330 + 41.8Tf + 627Tf – 15 675 = 0 Tf = 12 039
= 18.001°C 668.8
Ans: 18.00°C (b) Since the system is adiabatic, no heat enters or leaves the surroundings; therefore ∆Ssurr = 0, and ∆Suniv = ∆Ssyst = ∆S1 + ∆S2 + ∆S3 + ∆S4 = (10 × 2.04) ln "#
273.15$ 10 × 333
" 291.15 $
" 291.15 $
+
+ (10 × 4.18) ln #
+ (150 × 4.18) ln #
%
%
258.15
273.15
273.15
298.15 %
= 1.15220 + 12.19110 + 2.66772 – 14.89634 = + 1.11468 J K–1 Ans: + 1.115 J K–1 6. DuPont Canada uses acrylonitrile (C3H3N) as an important monomer for the production of many of their acrylic fibers. It can be synthesized from propene and ammonia according to the following reaction: 2 C3H6 (g) + 2 NH3 (g) + 3 O2 (g) → 2 C3N3N (l) + 6 H2O (g) Given that Δ So for this reaction is -­‐43.22 J K-­‐1 mol-­‐1, calculate the standard molar entropy (So) of acrylonitrile. 7. Problem 11–2 It has been proposed that methanol, which can be made from fast growing poplar trees, can be reacted according to the following reaction at 25°C and one bar pressure to provide a new source of natural gas (methane): methane gas, CH4 Δhof (25°C) (kJ mol–1) so2 9 8.15 (J K–1 mol–1) – 74.81 186.264 0.00 205.138 – 238.66 126.8 oxygen gas, O2 methanol liquid, CH3OH CH3OH(liq) → CH4(g) + 12 O2(g) Is this process feasible? Solution [ CH3OH(liq) → CH4(g) + 12 O2(g) The reaction is feasible if it is spontaneous; therefore, we must evaluate the minimum ∆Suniv. ∆Ssyst = ΔSoReact = so2 9 8.15 [CH4] + 12 so2 9 8.15 [O2] – so2 9 8.15 [CH3OH] = 186.264 + 12 (205.138) – 126.8 = + 162.033 J K–1 mol–1 For a constant pressure process, Qsyst = ∆Hsyst = ΔHoRe act = Δhof [CH4] + 12 Δhof [O2] – Δhof [CH3OH] = – 74.81 + 12 (0) – (– 238.66) = + 163.85 kJ mol–1 Also, Qsurr = Qsurr = – Qsyst = – 163.85 kJ mol–1 Therefore: ∆Ssurr = − 163 850
! Q#
= = – 549.556 J K–1 mol–1 " T $ surr
298.15
Finally: (ΔSuniv)min = ∆Ssyst + ∆Ssurr = + 162.033 – 549.556 = – 387.523 J K–1 mol–1 Since this is a negative value, the process is not spontaneous. The reaction is impossible. We can’t make methane using this reaction. Ans: No! ∆Suniv = – 387.5 J K–1 mol–1