Consider a system consisting of 100 gm of a liquid '4' present in an insulated beaker at 1 bar pressure. The liquid has been heated a heating coil operated by a 24 V battery, which provides

Question

Consider a system consisting of 100 gm of a liquid '4' present in an insulated beaker at 1 bar pressure. The liquid has been heated a heating coil operated by a 24 V battery, which provides a constant current of 2A, for 500 seconds under the influence of constant P ext ​ of 1 bar. During heating 80% of liquid has converted into vapours. Calculate change in internal energy of system, Given : density of A(l) = 1.4 g/ml density of A(g) = 0.6 g/mol 1 litre-bar = 700 J

TestHub · Accepted Answer

Heat supplied Q = V I t = 24 × 2 × 500 = 24000 J = 24 k J . Work done W = − P Δ V . Mass vaporized = 0.80 × 100 g = 80 g . Volume of liquid V l ​ = 1.4 g / m L 80 g ​ = 57.14 m L = 0.05714 L . Volume of vapor V g ​ = 0.6 g / m L 80 g ​ = 133.33 m L = 0.13333 L . Δ V = V g ​ − V l ​ = 0.13333 L − 0.05714 L = 0.07619 L . W = − 1 ba r × 0.07619 L = − 0.07619 L − ba r . Using 1 L − ba r = 700 J , W = − 0.07619 × 700 J = − 53.33 J . Change in internal energy Δ U = Q + W = 24000 J − 53.33 J = 23946.67 J ≈ 24 k J .

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