Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IWorkMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension

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

figure

What is the work done (in kJ) in going State 1 to State 3?

Options:

Answer:
B
Solution:

State 1 → 2: liquid is essentially incompressible, ΔV ≈ 0, so w ≈ 0. State 2 → 3: w = −Pext ΔV = −nRT = −25 × 8 × 373 = −74.6 kJ (work done on the gas; magnitude 74.6 kJ). Matches (b).

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Work
⏱ 2mℹ️ Source: QB

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