Review the worked solution for the problem below and identify the error$($s$)$ in the solution. For each error, identify the location of the error $($line #$),$ nature of the error $($numeric or conceptual$)$ and what the correct number$/$algebraic equation for that line should have been. Finally, provide the correct answer to the problem, if it had been worked correctly.

Air within a piston$-$cylinder assembly executes a Carnot heat pump cycle. For the cycle, $T_H=600K$ and $T_C=300K.$ The energy rejected by heat transfer at $600$ K has a magnitude of $250$ kJ per kg of air. The pressure at the start of the isothermal expansion is $325$ kPa $.$ Assuming the ideal gas model for air, determine $($a$)$ the magnitude of the net work input, in kJ per kg of air, and $($b$)$ the pressure at the end of the isothermal expansion, in kPa $.$

$(a)$ Because cycle:

$1,W_{\text{n}\text{e}\text{t}}=Q_{\text{n}\text{e}\text{t}}=Q_{in}-Q_{\text{o}\text{u}\text{t}}$

$2$ Whet $=Q_C-Q_H$

$(b)$ Because carnot:

$3,\frac{Q_c}{Q_H}=\frac{T_c}{T_H}$

$4,Q_c=Q_H(\frac{T_c}{T_H})=$

$5,=(250\frac{(kJ)}{(kg)})[\frac{300(K)}{600(K)}]$

$6,Q_c=125k\frac{J}{k}g$

$7$:$.w_{\text{n}\text{e}\text{t}}=(125-250k\frac{J}{k}g)$

$8,|W_{\text{n}\text{e}\text{t}}|=|-125k\frac{J}{k}g|$

$9,|w_{\text{n}\text{e}\text{t}}|=125k\frac{J}{k}g$

$(c)$ Find $P_3$ :

$*$ Because isothermal ideal gas: $U_{23}=0$:$.W_{23}=Q_{23}=-250\frac{(kJ)}{(kg)}$

$11,{}_{23}={\displaystyle \underset{v_2}{\overset{v_3}{}}}pdv={\displaystyle \underset{v_2}{\overset{v_3}{}}}\frac{RT}{v}dv=R{T}_{2}ln(\frac{v_3}{v_2})$

$12,\frac{v_3}{v_2}=e^{(\frac{W_{23}}{RT})v_2}=$exp$(\frac{-250\frac{kJ}{k}g}{(0\mathrm{.}287\frac{kJ}{k}g-k)(600k)})$

$13,\frac{v_3}{v_2}=0\mathrm{.}234$

$14,T_2=T_3$

$15,\frac{P_2}{B}$

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