Answer to Question #179095 in Mechanical Engineering for Danry

Question #179095

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.


1
Expert's answer
2021-04-12T02:33:01-0400

a)S = mCvln(T2/T1)

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

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

Q = mCp(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

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

b) U = mCvT = 1.36 kg x 306.75 J/kg K x (766.35 - 310.93 K) = 189 992 J = 190 kJ

W = Q - 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)


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