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Thermodynamics (Physics) · Question 12

NEET 2018MCQMedium📈 Topic in 25% papers+4−1
1

A sample of 0.1 g of water at 100°C and normal pressure (1.013 105 N m-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

  1. 104.3 J
  2. 208.7 J
  3. 42.2 J
  4. 84.5 J

Answer

(2) 208.7 J

Solution

Heat supplied: Q = 54 cal = 54 4.18 J = 225.72 J

Volume of 0.1 g of water = 0.1 cc

⇒ V = 167.1 - 0.1 = 167 cc = 167 10-6 m3

Work done by the sample: W = P V

= 1.013 105 167 10-6

= 16.92 J

First law of thermodynamics: U = Q - W

= 225.72 - 16.92

208.8 J 208.7 J

Answer: (2)

NEET 2018

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