Answer
$m_1 = 2.46\ kg$
Work Step by Step
For isentropic processes:
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^\dfrac{k-1}{k}$
Given $k=1.667,\ P_2=200\ kPa,\ P_1=450\ kPa,\ T_1=303\ K$
$T_2=219\ K$
$\dfrac{P_1V_1}{m_1RT_1}=\dfrac{P_2V_2}{m_2RT_2}$
$m_2=\dfrac{P_2T_1}{P_1T_2}m_1,\quad m_1=4\ kg$
$m_1 = 2.46\ kg$
