# 8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

Dec 7, 2014

The gas' new volume will be ${V}_{2} = 7.28$ L.

This problem is a simple application of Charles' law

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

which states that the volume of a gas is directly proportional to its temperature(measured in $K$ - very important).

So, we have an initial volume equal to ${V}_{1} = 8.00 L$, an initial temperature of ${T}_{1} = \left(273.15 + 60\right) = 333.15 K$, and a final temperature of ${T}_{2} = \left(273.15 + 30\right) = 303.15 K$, which gets us

${V}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1} = \frac{303.15}{333.15} \cdot 8.00 = 7.28 L$ -> a decrease in temperature is correlated with a decrease in volume, just as expected.

Note, however, what would have happened with degrees Celsius instead of K:

${V}_{2} = \frac{30}{60} \cdot 8.00 = 4.00 L$, a result significantly different...