Calculate the entropy change when a. two moles of H2O(g) are cooled irreversible at constant p from 120°C to 100°C. b. one mole of H2O(g) is expanded at constant pressure of 2 bar from an original volume of 20 L to a final volume of 25 L. You can consider the gas to be ideal. c. one hundred grams of H2O(s) at -10°C and 1 bar are heated to H2O(l) at +10°C and 1 bar.

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

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 = 0.00795 kJ/(K)

= 7.95 J/K

b.  At constant pressure for one  mole = 18 grams

Therefore change in entropy S₂ - S₁ =

= = -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 = = -0.03078 kJ/K

= -30.78 J/k

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