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Question
A sample of 0.1 g of water at 100°C and normal pressure (1.013×105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is
A sample of 0.1 g of water at 100°C and normal pressure (1.013×105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is
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Solution
The correct option is B 208.7 J
ΔQ = 54 cal = 54 × 4.18 J = 225.72 JΔW = P[Vsteam – Vwater] [For 0.1 gram of water, volume =0.1 cc] = 1.013 × 105 [167.1 × 10–6 – 0.1 × 10–6] J = 1.013 × 167 × 10–1 = 16.917 JBy First law of thermodynamics,ΔU =ΔQ – ΔW = 225.72 – 16.917 = 208.803 J
ΔQ = 54 cal = 54 × 4.18 J = 225.72 JΔW = P[Vsteam – Vwater] [For 0.1 gram of water, volume =0.1 cc] = 1.013 × 105 [167.1 × 10–6 – 0.1 × 10–6] J = 1.013 × 167 × 10–1 = 16.917 JBy First law of thermodynamics,ΔU =ΔQ – ΔW = 225.72 – 16.917 = 208.803 J
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