Answer
a) $m_e=6.618\ kg$
b) $m_e=6.551\ kg$
c) $m_e=6.524\ kg$
Work Step by Step
a) $m=Pv/RT$
Constant $V=1\ m³,\ R=0.287\ kJ/kg.K$
Initial $P_1=800\ kPa,\ T_1=25°C$: $m_1=9.354\ kg$
Final $P_2=150\ kPa,\ T_2=?$: $m_2=522.6/T_2$
Energy balance:
$-m_eh_e=m_2u_2-m_1u_1$
$-(m_1-m_2)\frac{c_p}2(T_1+T_2)=m_2c_vT_2-m_1c_vT_1$
$c_v=0.718\ kJ/kg.K,\ c_p=1.005\ kJ/kg.K,\ T_1=298\ K$
$T_2=191.0\ K$
$m_2=2.736\ kg$
$m_e=m_1-m_2=6.618\ kg$
b) Intermediary $P_2=400\ kPa,\ T_2=?$: $m_2=1394/T_2$
$-m_eh_e=m_2u_2-m_1u_1$
$-(m_1-m_2)\frac{c_p}2(T_1+T_2)=m_2c_vT_2-m_1c_vT_1$
$T_2=245.1\ K$
$m_2=5.687\ kg$
Final $P_3=150\ kPa,\ T_3=?$: $m_3=522.6/T_3$
$-m_eh_e=m_3u_3-m_2u_2$
$-(m_2-m_3)\frac{c_p}2(T_2+T_3)=m_3c_vT_3-m_2c_vT_2$
$T_3=186.5\ K$
$m_3=2.803\ kg$
Total $-m_e=m_1-m_3$
$m_e=6.551\ kg$
c) $\frac{dm}{dt}=-\dot{m}_e$
$\frac VR\frac{d(P/T)}{dt}=-\dot{m}_e$
$\frac{d(mu)}{dt}=-h_e\dot{m}_e$
$c_v\frac{d(mT)}{dt}=c_pT\frac{dm}{dt}$
$\frac VR \frac{dP}{dt}=kT\frac VR\frac{d(P/T)}{dt}$
$\frac{dP}{dt}=k[\frac{dP}{dt}-\frac PT \frac{dT}{dt}]$
$(k-1).\frac{d(\ln P)}{dt}=k.\frac{d(\ln T)}{dt}$
Integrating over time:
$\left (\dfrac{T_2}{T_1}\right)^k=\left(\dfrac{P_2}{P_1}\right)^{k-1}$
$T_2=184.7\ K$
$m_2=2.830\ kg$
$m_e=6.524\ kg$
