# Question #36619

Dec 31, 2017

Because its absolute temperature didn't double.

#### Explanation:

As you know, when the pressure and the number of moles of gas present in a container are kept constant, the volume of the gas and its temperature have a direct relationship as described by Charles' Law.

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

Here

• ${V}_{1}$, ${T}_{1}$ represent the volume and the absolute temperature of the gas at an initial state
• ${V}_{2}$, ${T}_{2}$ represent the volume and the absolute temperature of the gas at a final state

The key thing to keep in mind here is that the temperature of the gas must be expressed in Kelvin in order for this equation to work.

To do that, you use the conversion factor

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15}}}$

In your case, the temperature of the gas doubles in degrees Celsius, but that does not mean that it will double in Kelvin!

${T}_{1} = {25}^{\circ} \text{C" + 273.15 = "298 K}$

${T}_{2} = {50}^{\circ} \text{C" + 273.15 = "323 K}$

This means that the volume of the gas, which is equal to

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

will be equal to

${V}_{2} = \left(323 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{K"))))/(298color(red)(cancel(color(black)("K}}}}\right) \cdot {V}_{1}$

${V}_{2} \approx 1.084 \cdot {V}_{1}$

So the volume of the gas will only increase $1.084$ times when the temperature of the gas goes from ${25}^{\circ} \text{C}$ to ${50}^{\circ} \text{C}$.

In order for the volume to double, i.e. to have ${V}_{2} = 2 {V}_{1}$, the temperature of the gas must increase to

${T}_{2} = \frac{2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}} \cdot {T}_{1}$

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

${T}_{2} = \text{596 K}$

In degrees Celsius, the temperature of the gas must increase to

${t}_{2} = \text{596 K"- 273.15 = 323^@"C}$

in order for the volume of the gas to double.