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

Question

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 (A) Zero (B) 50 kJ (C) -50 kJ (D) 250 kJ

Tardigrade Admin · Accepted Answer

For a cyclic process, the work done is equal to heat absorbed q = Δ E - W ∴ Δ E 0 thus, q = -W = 50 kJ

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

Zero

50 kJ

-50 kJ

250 kJ

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

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

Zero

50 kJ

-50 kJ

250 kJ

For a cyclic process, the work done is equal to heat absorbed q=ΔE−W ∴ΔE0 thus, q = -W = 50 kJ

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