Answer
(a) The new volume is $1.27~V_0$
(b) The new volume is $2~V_0$
Work Step by Step
We can use the ideal gas law to solve this question.
$PV = nRT$
$V = \frac{nRT}{P}$
Let $V_0$ be the initial volume. Let $T_1$ be the initial temperature in Kelvin.
$V_0 = \frac{nRT_1}{P}$
(a) We can find the value of $T_1$ in Kelvin.
$K = C + 273$
$K = 100 + 273$
$K = 373$
If the Celsius temperature is doubled, we can find the new temperature $T_2$ in Kelvin.
$K = C + 273$
$K = 200 + 273$
$K = 473$
We can find an expression for $T_2$ in terms of $T_1$:
$\frac{T_2}{T_1} = \frac{473}{373}$
$T_2 = 1.27~T_1$
We can find an expression for the new volume $V_2$:
$V_2 = \frac{nRT_2}{P}$
$V_2 = \frac{nR(1.27~T_1)}{P}$
$V_2 = 1.27~\frac{nRT_1}{P}$
$V_2 = 1.27~V_0$
The new volume is $1.27~V_0$.
(b) If the Kelvin temperature is doubled, then $T_2 = 2~T_1$
We can find an expression for the new volume $V_2$:
$V_2 = \frac{nRT_2}{P}$
$V_2 = \frac{nR(2~T_1)}{P}$
$V_2 = 2~\frac{nRT_1}{P}$
$V_2 = 2~V_0$
The new volume is $2~V_0$.
