1. Tardigrade
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  3. Chemistry
  4. The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is -50 kJ/cycle. The heat absorbed by the systemper cycle is

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Q. The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is -50 kJ/cycle. The heat absorbed by the systemper cycle is

Thermodynamics

Solution:

For a cyclic process, the work done is equal to heat absorbed
$q = \Delta E - W$
$\therefore \, \Delta E 0$
thus, q = -W = 50 kJ