# If the volume of an ideal gas at 27^@ "C" halves, what is its new temperature? Why can I not use the celsius temperature to do this?

May 12, 2017

${330}^{\circ} \text{C}$

Assuming an ideal gas and that we are at constant pressure and mols of ideal gas, we construct two ideal gas states:

$P {V}_{1} = n R {T}_{1}$
$P {V}_{2} = n R {T}_{2}$

Solving for $\frac{n R}{P}$, we realize...

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

which is Charles law, relating volume and temperature changes at constant pressure and mols of gas .

Designate the original volume as $V$. Note that if we are sloppy and do not change temperature units, we have failed miserably.

${T}_{2} = \left({V}_{2} / {V}_{1}\right) {T}_{1}$

$= \frac{\cancel{V}}{0.5 \cancel{V}} \left({27}^{\circ} \text{C}\right)$

$\implies {T}_{2} = \underline{\textcolor{red}{{54}^{\circ} \text{C}}}$

which is completely incorrect... What we should have done is recognize that ${27}^{\circ} \text{C" + 273.15 -> "300.15 K}$. Hence, what we should have written is:

$\textcolor{b l u e}{{T}_{2}} = \frac{1}{0.5} \left(\text{300.15 K}\right)$

$=$ $\textcolor{b l u e}{{6.0}_{03} \times {10}^{2} \text{K}}$

or 327.15^@ "C" -> color(blue)(330^@"C") to two sig figs. That is, the KELVIN temperature is doubled, just as we expected when we wish to double the volume.

(Obviously, the celsius scale has destroyed the meaning of the logic; we do not see a clear-cut doubling of the celsius temperature.)