First law of Thermodynamics

Solved problems

1. For the melting of 1 mole of ice at 0oC and 1 atm are needed 1440 calories. By knowing that the molar volume of water and ice are 0.0180  and 0.0196 liter respectively, calculate DH and DE.

DH is 1440 cal and since DH = DE + PDV, in order to calculate DE we need to calculate PDV

PDV = 1 x (0.0180 - 0.0196) = -1.6 x 10-3 liter-atm

since 1 liter-atm is 24.4 cal PDV = -0.039 cal which is negligible when compared to DH thus DE = 1440 cal.

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2.  Calculate the DE for the following reaction:

C + 1/2 O2 = CO,   DH = -26416 cal at 298 K and 1 atm.

the molar volume of graphite is 0.0053 liter.

DH = DE + PDV, in order to calculate DE we need to calculate PDV. The variation of volume is due both to the disappearance of solid graphite and to the variation of number of moles (1 - 1/2 = 0.5). Remembering that a mole of a gas at standard condition (0 oC and 1 atm) occupies 24.4 liters:

DV = (0.5) (24.4) = 12.2 liter

Thus the variation of volume due to the variation of number of moles is 12.2 and, by neglecting the variation of volume due to the disappearance of solid graphite (0.0053 liter),

PDV = (1 atm)(12.2 liter) = 12.2 liter-atm

Since 1 liter-atm is 24.4 cal,  PDV = (12.2) (24.4) = 297.68 cal

DH = DE + PDV
DE = -26416 - 297.68 = -26713 cal

Alternatively and more simple, remembering that PDV = DnRT and that Dn = 0.5

PDV = 0.5 RT = (O.5)(1.987) (298) = 296
DE = -26416 - 296 = 26712 cal

In this last case, care must be taken in the choice of R (here cal/mole deg).

 
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3. In a furnace carbon is completely oxidized to CO2 by reaction with air, air and carbon enter the furnace at 60oF. The DH for the reaction of combustion is -172000 Btu/lb mole. Calculate the temperature reached in the furnace using the following table of specific heats:

 
oF
Cp N2
Cp CO2
1000
7.16
10.85
3000
7.85
12.76
5000
8.20
13.60

Furthermore, anticipate the effect on temperature when more air than theoretically required for the complete combustion is used.

The reaction to consider is

C + O2 + 3.7N2 = CO2 + 3.7N2

The energy balance can be written as

HCoke + Hair + Hreaction = HCO2 + HN2

Taking 60oF (the temperature at which coke and air enter the furnace) as reference temperature HCoke and Hair become zero thus

Hreaction = HCO2 + HN2 = 172000

172000 = 3.7Cp(N2) (Tf - 60) + Cp (CO2) (Tf - 60)

where Tf is the temperature reached in the furnace. Since Tf is unknown also the specific heats are unknown thus we should proceed by trials. Using the table given above we can assume a value for Tf and calculate the energy liberated from the reaction

oF
Cp N2
Cp CO2
Energyliberated
1000
7.16
10.85
35101
3000
7.85
12.76
122907
5000
8.20
13.60
217063

thus the value of Tf is between 3000 and 5000 oF and by linear interpolation we get Tf = 4042 oF

If more air than theoretical is admitted, the energy liberated from the reaction will be distributed on more material and then Tf will be lower.

 
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4. At pressure of 1 atm the enthalpy of evaporation of water is 9710 cal mole-1 and the variation of entropy is 26 cal mole-1 K-1. Calculate the temperature at which water is in equilibrium with its vapor (steam).

DG = DH - T DS = 9710 - 26 T

At equilibrium DG = 0 thus 9710 - 26 T = 0 from which

T = 9710/26 = 373 oK = 100 oC

Thus at 100 oC (the boiling point of water) water is in equilibrium with steam.