7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40   K at constant pressure.…

7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is
(Given R=8.3 J K-1 mol-1)
  1. 5810 J
  2. 3486 J
  3. 11620 J
  4. 6972 J

Solution

For the given process pressure is constant therefore, it is an isobaric process.

For a quasi-static process the change in internal energy of an ideal gas is independent of the nature of the process and is given by,

ΔU=nCvΔT

=n×3R2×ΔT

[molar heat capacity at constant volume for monatomic gas =3R2]

ΔU=7×32×8.3×40=3486 J

Asked in: JEE Main 2022 (26 Jul Shift 1)

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