Answer
(a) The temperature increases by a factor of 6
(b) The temperature increases by a factor of 1.5
Work Step by Step
We can solve this problem by reorganizing the ideal-gas law to solve for temperature:
$pV = nRT$
$T = \frac{pV}{nR}$
(a) $T_1 = \frac{p_1V_1}{nR}$
We can find an expression for the new temperature $T_2$, where $V_2 = 2V_1$ and $p_2 = 3p_1$:
$T_2 = \frac{p_2V_2}{nR}$
$T_2 = \frac{(3p_1)(2V_1)}{nR}$
$T_2 = 6~\frac{p_1V_1}{nR}$
$T_2 = 6~T_1$
The temperature increases by a factor of 6.
(b) $T_1 = \frac{p_1V_1}{nR}$
We can find an expression for the new temperature $T_2$, where $V_2 = \frac{1}{2}V_1$ and $p_2 = 3p_1$:
$T_2 = \frac{p_2V_2}{nR}$
$T_2 = \frac{(3p_1)(\frac{1}{2}V_1)}{nR}$
$T_2 = \frac{3}{2}~\frac{p_1V_1}{nR}$
$T_2 = \frac{3}{2}~T_1$
The pressure increases by a factor of $\frac{3}{2}$ or 1.5.
