Question #156517

For cretain ideal gas, R = 0.32 KJ/kg_{m }- K and c_{p} = 1160 J/kg_{m}- K. (a) Find c_{v} and k. (b) If 6 kg_{m} of this gas undergo a reversible nonflows constant pressure process from V_{1} = 2.1 cu. m, p_{1} = 0.7 MPaa to a state where T_{2} = 830 K, find â–²H, â–²U, and Q and W.

Expert's answer

(a) R = C_{p} - C_{v}

C_{v} = C_{p} - R = 1160 J/kgK - 320 J/kgK = 840 J/kgK

k = C_{p} / C_{v} = 1160 / 840 = 1.38

(b) m = 6 kg

p = const

V_{1} = 2.1 m^{3}

p = 0.7 MPa = 7 x 10^{5} Pa

T_{2} = 830 K

T_{1} = pV_{1} / mR = (7 x 10^{5} Pa × 2.1 m^{3}) / (6 kg x 320 J/kgK) = 765.6 K

V_{2} = T_{2}mR / p = (830 K × 6 kg x 320 J/kgK) / 7 x 10^{5} Pa = 2.28 m^{3}

Q = âˆ†H = mC_{p}(T_{2} - T_{1}) = 6 kg x 1160 J/kgK x (830 K - 765.6 K) = 448 224 J = 448 kJ

âˆ†U = mC_{v}(T_{2} - T_{1}) = 6 kg x 840 J/kgK x (830 K - 765.6 K) = 324 576 J = 325 kJ

W = pâˆ†V = p x (V_{2} - V_{1}) = 7 x 10^{5} Pa x (2.28 m^{3} - 2.1 m^{3}) = 126 000 J = 126 kJ

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