A container with a volume of 6 L contains a gas with a temperature of 720^o K. If the temperature of the gas changes to 420^o K without any change in pressure, what must the container's new volume be?

Oct 24, 2016

The new volume will be 4 L rounded to one significant figure.

Explanation:

This question involves the use of Charles' law, which states that the volume and temperature are directly proportional as long as pressure and amount are kept constant. This means that if the temperature increases, the volume increases and vice versa.

Given
${V}_{1} = \text{6 L}$
${T}_{1} = \text{720 K}$
${T}_{2} = \text{420 K}$

Unknown
${V}_{2}$

Solution
Rearrange the equation to isolate ${V}_{2}$. Substitute the given values into the equation and solve.

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

V_2=((6"L")*(420cancel"K"))/(720cancel"K")="4 L" rounded to one significant figure