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A 2 L vessel is filled with air 50 ° C and a pressure of 3 atm. The temperature is now raised to 200 ° C . A value is now opened so that the pressure inside drops to one atm. What will be the fraction of the total number of moles, inside escaped on opening the value? (Assume no change in the volume of the container).

Options

  1. A7.7
  2. B9.9
  3. C8.9
  4. D0.77

Correct answer

D. 0.77

Step-by-step solution

Given that, V = 2 L , T 1 = 50 + 273 = 323 K , P 1 = 3 a t m On heating V = 2 L , T 2 = 200 + 273 = 473 , P 2 = ? Using P-T law, P 1 T 1 = P 2 T 2 P 2 = 3 × 473 323 = 4.39 atm and n = P V R T = 3 × 2 0.0821 × 323 = 0.226 Now value is opened till the pressure is maintained at 1 atm. Thus, at constant V and T, P ∝ n So 4.39 ∝ 0.226 So 1 ∝ n l e f t So n = 0.226 4.39 = 0.052 So moles escaped out = 0.226 - 0.052 = 0.174 So fraction of moles escaped out = 0.174 0.226 = 0.77

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