Skip to content

Question

In an isobaric process, 0.9 moles of gas expand from 360 K to 450 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻¹ ?

Options

Choose one · Correct answer highlighted

Explanation

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. Δ Q = μ C_p Δ T . μ = 0.9 , C_p = 25.5 , Δ T = 450 - 360 = 90 . Δ Q = 0.9 × 25.5 × 90 = 2065.5 J ≈ 2066 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ

Discussion

Comments

0 comments

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