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in Chemical thermodynamics by (64.8k points)

The vapor pressure of H2O (l) 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 33 K is increased to 1 atm. Temperature remained constant. Now, heat is supplied keeping pressure constant till 0.45 L of H2O (l) (=0.45 kg) is evaporated to form H2O (g) at 373 K. carefully observe the diagrams provided and form given data, answer the following questions Given:

Specific heat of H2O = 4.2J/gm⁰C

Use Hvap at 373 K and 1 atm =+40 kJ/mol

1L atm = 100 Joule

1 atm = 760 mm Hg

R = 8 Joule/mole K 

(Assume internal energy of liquid to be dependent only on temperature).

ΔH When system is taken from state 1 to state 2 (Joule)? 

(A) Zero 

(B) 0.27 

(C) 27 

(D) 90 

1 Answer

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Best answer

Correct option (C) 27

Explanation:

(i) T = 532 mm Hg = 0.4 atm T = 353 k

Vi = 0.4 = 0.4 kg

∆H = ∆U + ∆pV

Dependent only on temperature

∴ ∆H = ∆pV = (1–0.7) × 0.9 L

= 0.3 × 0.9 L = 0.3 × 0.9 × 100 J= 27 J

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