Answer
(a) $983^{\circ}C$
(b) $0.0894 g$
Work Step by Step
For (a):
From the ideal gas law, $pV = nRT$. Here, $n$ and $R$ are constant, so $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$.
Note that $T_1 = 41^{\circ}C = 314 K$.
$T_2 = T_1 (\frac{p_2}{p_1})(\frac{V_2}{V_1}) = (314)(2)(2) = 1256 K = 983^{\circ}C$.
For (b):
$n = \frac{pV}{RT} = \frac{(0.180)(3.2)}{(0.082)(314)} = 0.02235\ mol$.
Mass $m = (0.02235)(4) = 0.0894 g$.
