3.5 moles of steam at 100 C are condensed at 100 C and the resulting water is cooled to 42.3 C. Calculate the entropy change for this process.(The molar heat capacity of liquid water is 75.3 J mol1 K1, and the molar enthalpy of vaporisation of water is 40.8 kJ mol/1 at 100 C).
10 J
To calculate the entropy change for this process, we need to consider the entropy changes during the condensation of steam at 100°C and the cooling of the resulting water to 42.3°C.
First, let's calculate the entropy change during the condensation of the steam.
The molar enthalpy of vaporization of water at 100°C is given as 40.8 kJ/mol. Entropy change during phase transition (condensation or vaporization) can be calculated using the formula:
ΔS = ΔH / T
where ΔS is the entropy change, ΔH is the enthalpy change, and T is the temperature.
Given:
ΔH = 40.8 kJ/mol
T = 100°C = 373.15 K
Converting ΔH to J/mol:
ΔH = 40.8 kJ/mol × 1000 J/1 kJ ≈ 40,800 J/mol
Using the formula to calculate ΔS:
ΔS = 40,800 J/mol / 373.15 K ≈ 109.53 J/(mol·K)
Next, let's calculate the entropy change during the cooling of the water from 100°C to 42.3°C.
For this process, we can use the heat capacity formula:
ΔS = n × C × ln(T2/T1)
where ΔS is the entropy change, n is the number of moles, C is the heat capacity, and ln is the natural logarithm.
Given:
n = 3.5 moles
C = 75.3 J/(mol·K)
T1 = 100°C = 373.15 K
T2 = 42.3°C = 315.45 K
Using the formula to calculate ΔS:
ΔS = 3.5 mol × 75.3 J/(mol·K) × ln(315.45 K / 373.15 K)
Using a calculator, we find:
ΔS ≈ -269.40 J/K
Finally, to find the total entropy change, we can sum the entropy changes from both processes:
Total entropy change = ΔS(condensation) + ΔS(cooling)
Total entropy change ≈ 109.53 J/(mol·K) + (-269.40 J/K)
Using a calculator, we find the total entropy change is approximately -159.87 J/(mol·K).