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Question 5 (2.0 pts) Assume that Cp for an ideal gas is given by co(T) = 0.9 + 2.5 x 10-* T, in kJ/kg-K, where is the tempera
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Isochoric process is also called as constant volume process.In constant volume process workdone is zero.so,according to first law of thermodynamics,heat transfer will be equal to change in internal energy.(Q=\DeltaU).

GIVEN DATA : Cp (T) = 0.9 +2.5*10*7 , in kJ kg.k R=0.3 KJ kg.k Solution: The gas undergoes an Isochoric process from a temper(b) Enthalpy, in kJ 72 DH = 16. dt is formula. 500 | (0.9 +2.5x1047) .dt = [0.25 +2.52102 =[6.44560 + 25 205 * 50g] – (0.9448(C) entropy, in KI kg.k We know, first law of Thermodynamics is. Qing out wi-2 ->U) divide 0) by T orm where, we know, for is500 = 16.6 +2.5x1045), TT - - - 500 (946 + 25206T .47 400 for understanding 500 Coob + 2.5810*) .dt It.dt= int 900 500 = [ows(d) change in internal energy. Du= co.dt a formula Ti (6.6 +2.5x1097) .IT 500 (0.8+ 2.581047) odt 500 - Forst +2.5**) = [0.62

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