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JEE Main201910 Jan 2019Evening ShiftPhysicsThermodynamicsActual

Half mole of an ideal monoatomic gas is heated at a constant pressure of 1   atm from 20 ° C to 90 ° C . Work done by the gas is ( Gas constant, R = 8.21 J mol - 1 K - 1 )

Options

  1. A73   J
  2. B581   J
  3. C291   J
  4. D146   J

Correct answer

C. 291   J

Step-by-step solution

Recall the formula of work done in terms of change in volume and pressure, in the case of isobaric process, where pressure is constant, W = P V 2 - P V 1 , now take the idea of ideal gas equation, W = n R T 2 - n R T 1 = 1 2 × 8.31 70 = 291   J .

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