Answer
$T_2=257\ K$
$P_2=956\ kPa$
Work Step by Step
From the material balance:
$-m_e=m_2-m_1$
Given $m_2=m_1/2$
$m_e=m_1/2$
From the energy balance:
$-m_eh_e=m_2u_2-m_1u_1$
$-h_e=u_2-2u_1$
$-c_pT_e=c_vT_2-2c_vT_1$
Assuming $T_e\approx\dfrac{T_1+T_2}{2}$:
$-k(T_1+T_2)=2T_2-4T_1$
Given $T_1=403\ K,\ k=1.667$:
and solving for the final temperature: $T_2=257\ K$
For ideal gases:
$PV=mRT$, and the volume is constant:
$\dfrac{m_1T_1}{P_1}=\dfrac{m_2T_2}{P_2}=\dfrac{m_1T_2}{2P_2}$
Given $P_1=3\ MPa$
$P_2=956\ kPa$
