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Home , Molar heat capacity

University of Washington Department of Chemistry 452/456 Summer Quarter 2010

Homework Assignment #3: Due at 5 p.m. on 7/12/10 Text: 5.7, 5.13, 5.18, 5.20

P.513) a) q=0 because the process is adiabatic

P5.18) kJ !HfGlycine := "537.2 # mol kJ !HfGg := "746.0 # mol kJ !HfH2O := "285.8 # mol joule SGlycine := 103.5 ! mol ! K joule SGg := 190.0 ! mol ! K joule SH2O := 70.0 ! mol ! K

!Ssys := SGg + SH2O # 2 " SGlycine -1 -1 !Ssys = 53.00 K mol joule

!H := !HfGg + !HfH2O # 2 " !HfGlycine -1 !H = 42.60 mol kJ The process is endothermic, and the surroundings lose this much . "!H !Ssurr := 298 # K -1 -1 !Ssurr = "142.95 K mol joule

!Suniverse := !Ssys + !Ssurr -1 -1 !Suniverse = "89.95 K mol joule

Problem 5.20

6 CO2(g) + 6 H2O(l) = C6H12O6(s) + 6 O2(g) kJ !HfCO2 := "393.5 # mol kJ !HfH2O := "285.8 # mol kJ !HfC6H12O6 := "1273.1 # mol joule SCO2 := 213.8 ! mol ! K joule SH2O := 70.0 ! mol ! K joule SC6H12O6 := 209.2 ! mol ! K joule SO2 := 205.2 ! mol ! K joule CpCO2 := 37.1 ! mol ! K joule CpH2O := 75.3 ! mol ! K joule CpH2O := 219.2 ! mol ! K joule CpO2 := 29.4 ! mol ! K !H298 := !HfC6H12O6 # 6 " !HfCO2 # 6 " !HfH2O -1 !H298 = 2802.7 mol kJ

!Cp := CpC6H12O6 + 6 " CpO2 # 6 " CpCO2 # 6 " CpH2O -1 -1 !Cp = "1.14 K mol kJ

!H330 := !H298 + !Cp " (330 " K # 298 " K) -1 !H330 = 2766.1 mol kJ

!S298 := SC6H12O6 + 6 " SO2 # 6 " SCO2 # 6 " SH2O -1 -1 !S298 = "262.40 K mol joule 330"K # $ !Cp !S330 := !S298 + dT $ T % 298"K -1 -1 !S330 = "379.01 K mol joule For the surroundings and the universe: The process is endothermic, so the surroundings give up heat.

"!H298 !Ssurr298 := 298 # K 3 -1 -1 !Ssurr298 = "9.41 # 10 K mol joule

!Suniverse298 := !S298 + !Ssurr298 3 -1 -1 !Suniverse298 = "9.67 # 10 K mol joule

"!H330 !Ssurr330 := 330 # K 3 -1 -1 !Ssurr330 = "8.38 # 10 K mol joule

!Suniverse330 := !S330 + !Ssurr330 3 -1 -1 !Suniverse330 = "8.76 # 10 K mol joule

1) Calculate ΔSsys ,ΔSsurr , and ΔSuniverse for the vaporization of one of superheated liquid at a of 383 K and a of 1 atm. At the normal boiling point T=373 K the of water is ΔHvap=40,700 J/mole. For liquid -1 -1 water the molar at constant pressure CP=75 J mole K . For water vapor -1 -1 the molar heat capacity at constant pressure is CP=33.5 J mole K . Hint: Describe a three step path. For each step calculate ΔS and ΔH. Assume all heat capacities are constant from 373-383 K.

Solution: liquid (383K ) ! vapor (383K ) " # liquid (373K ) ! vapor (373K ) H H nC liquid T H nC vapor T $ = $ sys = P ( )$ 373%383 + $ vap + P ( )$ 383%373 H n C T 40,700J 33.5J / K 75J / K 383K 373K = $ vap + $ P$ = + ( % )( % ) = 40,700J + (%41.5J / K )(10K ) = 40,700J % 415J = 40,285J & 373) $H & 383) $S = nC liquid ln + vap + nC vapor ln sys P ( ) '( 383*+ 373K P ( ) '( 373*+ 40,700J = 75J / K ln 0.974 + + 33.5J / K ln 1.03 ( ) ( ) 373K ( ) ( ) = %1.98J / K + 109J / K + 0.886J / K = 108J / K Then H !Hsurr ! sys 40285J !Ssurr = = " = " = "105J / K Tsurr Tsurr 383K !S = !S + !S = 108J / K " 105J / K = 3J / K universe sys surr 2) Calculate the entropy change if 2 moles of liquid at P= 1atm and T=233.2K are heated to T= 473K. The normal boiling point of ammonia is -1 239.7K. The heat of vaporization of ammonia is ΔHvap =23.2kJmol . The heat -1 -1 capacity of liquid ammonia is CP =74.8JK mol and is constant from 233K to 239.7K. The heat capacity of ammonia between 239.7k and 473K is given 2 "4 # T & "5 # T & -1 -1 by: C = 33.6 + 29.3 ! 10 + 21.3 ! 10 JK mol p $% K '( $% K '( Solution: 239.7 K C(!) !H 473K C(v) !S = " P dT + vap + " P dT 233.2 K T Tvap 239.7 K T #1 239.7 K 74.8JK #1mol #1 23200Jmol #1 473K $ $ T ' $ T ' ' = dT + + & 33.6 + 29.3* 10#4 + 21.3* 10#5 )JK #1mol #1dT " T 239.7K " & & K ) & K ) ) 233.2 K 239.7 K % % ( % ( ( $ 239.7' $ 473 ' = 74.8JK #1mol #1 ln + 96.8JK #1mol #1 + 33.6JK #1mol #1 * ln ( ) %& 233.2() %& 239.7() 2 2 #4 #1 #1 #5 #1 #1 $ 473 # 239.7 ' +29.3* 10 JK mol * (473# 239.7) + 21.3* 10 JK mol & ) % 2 ( = 2.06JK #1mol #1 + 96.8JK #1mol #1 + 22.8JK #1mol #1 + 0.68JK #1mol #1 + 17.7JK #1mol #1 = 140JK #1mol #1 + !S = 2mol 140JK #1mol #1 = 280JK #1 ( )( )

3) An ideal diatomic gas expands irreversibly and adiabatically from an initial state of V1=20L, P1=100atm, and T1=473K against a constant external pressure of 1 atm until the are balanced. Calculate the final volume, pressure, and -1 -1 temperature. Calculate the entropy change of the gas. Assume CP=29.3JK mol .

Solution:

P2 = 1atm adiabatic...q = 0 ! "U = nCv "T = w = #Pext "V #1 ( nC P + V $ nC T P V & P v 2 ! 2 = %# v 1 # ext 1 '* # ext # - ) nR , #1 ( 3 + 1mol R 1atm $ 3 &* ( ) ( )- V = # 1mol R 473K # 1atm 20L # 1atm # 2 = 31.3L 2 . ( ) 2 ( ) ( )( )/* ( ) - % '* (1mol) R - ) , 1atm 31.3L P2V2 ( )( ) T2 = = = 381K nR 1mol 0.0821L ! atm ! mol "1 ! K "1 ( )( ) # P & 381K nC # 100& 381K 29.3J * K "1 # 381& !S = "nRln 2 + P dT = 8.314ln J + dT = 38.3+ 29.3ln = 32J % P ( ) T $% 1 '( ) T $% 473'( $ 1 ' 473K 473K

4) From the data below, derive the absolute entropy of crystalline glycine at T=298K. Note the heat capacity is in antiquated units…calories per K per mole. There are 4.18J/cal. Give your answer in units of J/Kmole. Hint: Do the necessary calculus graphically, using the fact that an integral is an area under a curve. I want to see a graph of the data from which you get the necessary area. How you get the area accurately is up to you to determine.

Heat Capacity of Crystalline Glycine as a Function of Temperature.

Temperature (K) Heat Capacity Cp (cal/Kmole) 10 0.061 20 0.572 30 1.682 40 3.108 60 6.012 80 8.427 100 10.34 120 11.95 140 13.40 160 14.74 180 16.02 200 17.28 220 18.52 240 19.80 260 21.14 280 22.48 300 23.84

Solution: Plot Cp/T versus T…The entropy is the area under the curve which is ΔS=26.05 cal/K=108.9J/K.

Cp/T vs. T for Glycine

0.12

0.1

0.08

0.06

Cp/T (cal/T^2) 0.04

0.02

0 0 50 100 150 200 250 300 350 Temperature (K)