Question

One kg of air (R = 287 J/kg.K) undergoes an irreversible process between equilibrium state 1 (20° C, 0.9m³) and equilibrium state 2 (20°C, 0.6 m³). The change in entropy s₂ – s₁ (in J/kg.K) is _________

Answer 1

Concept:

From T-dS relation,

\(Tds=dh-vdp\)

\(Tds=C_p dT-vdp\)

for an isothermal process,

\({s_2} - {s_1} = R\ln \frac{{{V_2}}}{{{V_1}}}\)\( = R\ln \frac{{{P_1}}}{{{P_2}}}\)

Calculation:

Given, T₁ = 20^o C = 293 K, V₁ = 0.9 m³

T₂ =  20o C = 293 K, V₂ = 0.6 m³

\({s_2} - {s_1} = R\ln \frac{{{V_2}}}{{{V_1}}}\)

\(s_2 -s_1 = 287\ln \frac{{0.6}}{{0.9}} \)

\(s_2 -s_1 = -116.368 \) J/kg-K