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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).

Total change in ∆U going from state 1 to 3 (kJ)? 

(A) 75.6 

(B) 1075.6 

(C) 1001 

(D) 74.6

1 Answer

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Best answer

Correct option (B) 1075.6 

Explanation:

From 1 to 3

∆U = ∆ (nCT)

∆mCT = 0.9 × C × (373 – 353) + 0.4 x 40/18 = 1075.6

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