Answer
(a) $975K$
(b) $4atm$
(c) $5\times 10^{-4}m^3$
Work Step by Step
(a) We can find $T_1$ as follows:
$T_1=\frac{p_1V_1}{nR}$
We plug in the known values to obtain:
$T_1=\frac{1(101300)(1\times 10^{-4})}{(0.005)(8.31)}$
$T_1=975K$
(b) We know that
$p_2=\frac{nRT_2}{V_2}$
We plug in the known values to obtain:
$p_2=\frac{(0.005)(8.31)(2926)}{3\times 10^{-4}}$
This simplifies to:
$p_2=405251pa=4atm$
(c) We can find the volume $V_3$ as follows:
$V_3=\frac{nRT_3}{p_3}$
We plug in the known values to obtain:
$V_3=\frac{(0.005)(8.31)(2438)}{2(101300)}$
$V_3=5\times 10^{-4}m^3$
