Answer
$T_2=241\ K=-32°C$
Work Step by Step
For ideal gases:
$PV=mRT$
Given $V=0.5\ m³,\ R=0.287\ kJ/kg.K$
Initial $P_1=4\ MPa,\ T_1=20°C$: $m_1=23.78\ kg$
Final $P_2=2\ MPa,\ T_2=?$: $m_2=3484/T_2$
From the material balance:
$-m_e=m_2-m_1$
From the energy balance:
$-m_eh_e=m_2u_2-m_1u_1$
$(m_2-m_1)c_pT_e=m_2c_vT_2-m_1c_vT_1$
$T_e=\dfrac{T_1+T_2}2$
$\frac k2 (m_2T_2+m_2T_1-m_1T_2-m_1T_1)=m_2T_2-m_1T_1$
$\frac k2 (3484(1+\frac{T_1}{T_2})-m_1(T_2+T_1)))=3484-m_1T_1$
Given $k=1.4,\ T_1=293\ K$:
and solving for $T_2=241\ K=-32°C$
