Chemistry - Thermodynamics - I Question with Solution | TestHub
The vapour pressure of water at 353 K is 532 mm Hg. The external pressure on H₂O (l) taken in a cylinder fitted with frictionless movable piston initially containing 0.9 L (= 0.9 kg) of H₂O (l) at 353 K is increased to 1 atm at the constant temperature. Then, heat is supplied keeping the pressure constant till 0.45 L of H₂O (l) is evaporated to form H₂O (g) at 373 K. Assume the internal energy of liquid to be dependent only on temperature. Answer the following questions by carefully observing the diagrams and the data provided. Given:
Specific heat of H₂O (l) = 4.2 J/°C-g
ΔHvap at 373 K and 1 atm = +40 kJ/mol
1 L-atm = 100 J
R = 8 J/K-mol
Total change in ΔU (in kJ) going from State 1 to State 3 is
Options:
Answer:
Solution:
State 1 → 2: isothermal, ΔU = 0. State 2 → 3 (constant P = 1 atm): heat the 0.9 kg liquid from 353 K to 373 K, q₁ = 900 g × 4.2 × 20 = 75.6 kJ; evaporate 0.45 kg, n = 450/18 = 25 mol, q₂ = 25 × 40 = 1000 kJ. qp = ΔH = 1075.6 kJ. Work: w = −Pext ΔV = −P·Vgas = −nRT = −25 × 8 × 373 J = −74.6 kJ. ΔU = q + w = 1075.6 − 74.6 = 1001 kJ. Matches (c).
