Answer to Question #179095 in Mechanical Engineering for Danry

Question

An ideal gas whose mass is 1.36 kg m , R = 204.51 J/kg m and k = 1.6667. Heat added during a reversible nonflow constant pressure change of state is 316.5 kJ . The initial temperature is 37.78 o C. Determine a) change in entropy (kJ/K), b) Work (kJ), c) ratio of expansion or V 2 /V 1.

Answer 1

a)$\Delta$S = mC_vln(T₂/T₁)

C_v = R/(k - 1) = 204.51 J/kg K / (1.6667 - 1) = 306.75 J/kg K

C_p = kR/(k - 1) = (1.6667 x 0.20451 kJ/kg K) / (1.6667 - 1) = 0.511 kJ/kg K

Q = mC_p(T₂ - T₁)

316.5 kJ = 1.36 kg x 0.511 kJ/kg K x (T₂ - (37.78 + 273.15) K)

T₂ = 766.35 K

$\Delta$S = 1.36 kg x 0.30675 kJ/kg K x ln(766.35 K/310.98 K) = 0.376 kJ/K

b) $\Delta$U = mC_v$\Delta$T = 1.36 kg x 306.75 J/kg K x (766.35 - 310.93 K) = 189 992 J = 190 kJ

W = Q - $\Delta$U = 316.5 kJ - 190 kJ = 126.5 kJ

c) W = p₁V₁ln(V₂/V₁) = nRT₁ln(V₂/V₁)

126 500 J = (1360 g/Mr g/mol) x 8.31 J/mol K x 310.98 K x ln(V₂/V₁)