Answer to Question #274655 in Molecular Physics | Thermodynamics for note

Question

An ideal gas with Cv = 1.62 kJ/kg-K and R = 273J/kg-K expands from 0.15 m3 and 27oC to 0.35 m3 during an isobaric process with p = 150kPa. Consider the process reversible nonflow, determine (a) the heat, (b) the change in internal energy, © the work nonflow, (d) the change in entropy. 

Answer 1

a)

$T_1=300~K,$

$T_2=T_1\frac{p_2}{p_1}=700~K,$

$Q=\nu R\Delta T=53.2~kJ,$

b)

$\Delta U=p\Delta V=30~kJ,$

c)

$A=Q-\Delta U=23~kJ,$

d)

$S=\frac Q{\Delta T}=133~\frac JK.$