Respuesta :
Answer:
The answers to the question are
a. S₂ - S₁ = 7.95 J/K
b. S₂ - S₁ = -16.9 J/K
c. S₂ - S₁ = -30.78 J/k
Explanation:
a. To solve the question, we note that
The change in entropy S₂ - S₁ is given by
[tex]mC_pln(\frac{T_1}{T_2} )[/tex] Where
T₁ = Initial temperature in Kelvin = 120 + 273.15 K = 393.15 K
T₂ = Final temperature in Kelvin = 100 + 273.15 K = 373.15 K
and Cp for water = 4.23 kJ/(kg·K)
Molar mass of water = 18 g/mol
Therefore mass of water = 2*18 =36 grams = 0.036 kg
The change in entropy then is [tex]0.036*4.23*ln(\frac{393.15}{373.15} )[/tex] = 0.00795 kJ/(K)
= 7.95 J/K
b. At constant pressure for one mole = 18 grams
[tex]\frac{T_1}{T_2} = \frac{V_1}{V_2}[/tex] Therefore change in entropy S₂ - S₁ =[tex]mC_pln(\frac{T_1}{T_2} ) = mC_pln(\frac{V_1}{V_2} )[/tex]
= [tex]0.018*4.2*ln(\frac{20}{25} )[/tex] = -0.01686965 kJ/(K) = -16.9 J/K
c. 100 grams of H₂O from -10 °C (263.15 K) to 10 °C (283.15)
100 g = 0.1 kg
we have S₂ - S₁ is given by [tex]mC_pln(\frac{T_1}{T_2} )[/tex] = [tex]0.1*4.2*ln(\frac{263.15}{283.15} )[/tex] = -0.03078 kJ/K
= -30.78 J/k
