A container with a volume of 12 L contains a gas with a temperature of 230^o C. If the temperature of the gas changes to 420 ^o K without any change in pressure, what must the container's new volume be?

Mar 18, 2018

I get approximately $10 \setminus \text{L}$.

Explanation:

We first convert ${230}^{\circ} \text{C}$ into $\text{K}$, and we have:

${230}^{\circ} \text{C"=(273.15+230) \ "K}$ (since $\text{K"=""^@"C} + 273.15$)

$= 503.15 \setminus \text{K}$

If there is no change in pressure and the number of moles, we may use Charles's law, which states that

$V \propto T$

or

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

We need to solve for the new volume, so we rearrange the equation in terms of ${V}_{2}$, and we get:

${V}_{2} = {V}_{1} / {T}_{1} \cdot {T}_{2}$

Plugging in the given values, we find that

V_2=(12 \ "L")/(503.15color(red)cancelcolor(black)"K")*420color(red)cancelcolor(black)"K"

$\approx 10 \setminus \text{L}$ to the nearest whole number.