# What steps and formulas would you use to solve the question "At what Kelvin temperature will a sample of gas occupy 12 liters if the same sample occupies 8 liters at 27 °C?"

Feb 24, 2015

You'd use Charles' Law, which links an ideal gas' volume to its temperature if pressure and number of moles are kept constant.

The mathematical formula is this

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

${V}_{1} = 8.0 L$,
${T}_{1} = 273.15 + 27 = \text{300.15 K}$, and
${V}_{2} = \text{12.0 L}$

Therefore,

${T}_{2} = {V}_{2} / {V}_{1} \cdot {T}_{1} = \text{12.0 L"/"8.0 L" * "300.15 K" = "450 K}$

If pressure and number of moles are constant, an increase in volume means an increase in temperature and vice versa.

SIDE NOTE All you really need to know is the ideal gas law equation, PV = nRT, since you can apply it to all sorts of conditions; in your case, you'd get two equations to describe the two stages - once again, number of moles and pressure are constant

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

DIvide these two equations and you'll get the expression for Charles' law again

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

Feb 24, 2015

$V \propto T$

So:

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

You need to convert to Kelvin degrees by adding 273:

${V}_{1} = 8 L$
${T}_{1} = 27 + 273 = 300 K$
${V}_{2} = 12 L$
${T}_{2}$ is unknown.

${T}_{2} = \frac{{V}_{2} \times {T}_{1}}{{V}_{1}}$

${T}_{2} = \frac{12 \times 300}{8} = 450 K$