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Example-1

Example-1
About 10 L of a monoatomic ideal gas at 0°C and 10 atm pressure is suddenly released to 1 atm pressure and the gas expands adiabatically against this constant pressure. The final temperature and volume of the gas, respectively, are:
  1. T = 174.9 K, V = 64.04 L
  2. T = 153 K, V = 57 L
  3. T = 165.4 K, V = 78.8 L
  4. T = 161.2 K, V = 68.3 L
Solution (A)
This is an adiabatic reversible process. So for this process,
 
PVy = Constant is not applicable
 
w = –Pext(V2 – V1)
 
But for adiabatic process,
w = du = Description: 45530.png
 
PV = nRT = 10 × 10 = n × 0.082 × 273
 
Pext(V2 – V1) = Description: 45537.png
 
– 1 × (V2 – 10) = Description: 45545.png
 
Description: 45554.png
 
6.7 – 0.67V2 = V2 – 100
 
106.7 = 1.67V2
 
V2 = 64
 
nR[T2 – T1] = P2V2 – P1V1
 
4.47[0.082] [T2 – 273] = 64 – 100 = – 36
 
(T2 – 273) = –98.2
 
T2 = 174.8
 

Example-2

Example-2
The increase in internal energy of 1 kg of water at 100°C when it is converted into steam at the same temperature and at 1 atm (100 kPa) will be [the densities of water and steam are 1000 and 0.6 kg/m3, respectively. The latent heat of vaporization of water is 2.25 × 106 J/kg]:
  1. 2.08 × 106 J
  2. 4 × 107 J
  3. 3.27 × 108 J
  4. 5 × 109 J
Solution (A)
Latent heat of vaporization of water = 2.25 × 106 J/kg
 
ΔH = 2.25 × 106 J/kg
 
Work done = –Pext(V2 – V1)
 
ΔH = 2.25 × 106 J/kg
 
ΔH = ΔU + PΔV
  1. Now volume of water
     
    Description: 45566.png
     
    Description: 45575.png
  2. Volume of steam
     
    Description: 45583.png Description: 45592.png
     
    2.25 × 106 = ΔU + 1[1666.87 –1] 101.325
     
    ΔU = 22.5 × 105 – 1.68 × 105 = 20.8 × 105
     
    ΔU = 2.08 × 106 J

Example-3

Example-3
The heat of formation of C2H5OH(l) is –66 kcal/mol. The heat of combustion of CH3OCH3(g) is –348 kcal/mol. ΔHf for H2O and CO2 are –68 kcal/mol and –94 kcal/mol, respectively. Then ΔH for the isomerization reaction C2H5OH(l) → CH3OCH3(g), and ΔE for the same are:
  1. ΔH = 18 kcal/mol, ΔE = 17.301 kcal/mol
  2. ΔH = 95 kcal/mol, ΔE = 100.3 kJ/mol
  3. ΔH = 26 kcal/mol, ΔE = 25.709 kcal/mol
  4. ΔH = 30 kcal/mol, ΔE = 28.522 kcal/mol
Solution (B)

ΔH5 = 3 Eq. (iii) + 2 Eq. (ii) – Eq. (i) – Eq. (ii)
 
= + 66 + 348 – 3 × 68 – 2 × 94 + 66 + 348 – 204 – 188
 
ΔH = 22 kcal/mol
 
ΔH = ΔE + ΔngRT
 
Δng = 1 kcal/mol
 
22000 = ΔE + 1 × 2 × 298
 
ΔE = 21.408 kcal/mol
 

Example-4

Example-4
In the reaction:
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ΔH = –265 kcal
 
The enthalpies of formation for both CO2 and SO2 are negative and are in the ratio 4 : 3. The enthalpy of formation for CS2 is +26 kcal/mol. Calculate the enthalpy of formation for SO2:
  1. –90 kcal/mol
  2. –52 kcal/mol
  3. –78 kcal/mol
  4. –71.7 kcal/mol
Solution (D)
S2(l) + 3O2(g) → CO2(g) + 2SO2(g)
 
ΔH = –265 kcal
 
ΔHF° (CO2, g) = 4x
 
ΔHF° (SO2, g) = 3x
 
ΔHF° = ΔHF° (CO2, g) + 2ΔHF° (SO2, g) –ΔHF° (CS2, g)
 
–265 = 4x + 6x – 26
 
10x = 239
 
x = – 23.9
 
ΔHF° (SO2, g) = – 71.7 kcal/mol
 

Example-5

Example-5
The standard enthalpy of formation of FeO and Fe2O3 is –65 kcal/mol and –197 kcal/mol, respectively. A mixture of two oxides contains FeO and Fe2O3 in the mole ratio 2 : 1. If by oxidation, it is changed into a 1 : 2 mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture?
  1. 13.4 kcal/mol
  2. 14.6 kcal/mol
  3. 15.7 kcal/mol
  4. 16.8 kcal/mol
Solution (A)
Δ H3 = Eq. (iii) – 2 Eq. (i) = –67 kcal
Description: 45667.png
Description: 45676.png
So energy released = Description: 45744.png = 13.4 kcal/mol
 




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