Skip to content

Question

A system in a cyclic process absorbs 900 J of heat and rejects 400 J . What is the net work done?

Options

Choose one · Correct answer highlighted

Explanation

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For cyclic process: Δ U = 0 , W = Q_net . Q_net = Q_absorb - Q_reject = 900 - 400 = 500 J . W = 500 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.