# A 20.0 mL sample of a gas is at 546 K and has a pressure of 6.0 atm. If the temperature is changed to 273 K and the pressure to 2.0 atm, the new volume of the gas is what?

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, given constant $n$, so............
${V}_{2} = \frac{{P}_{1} \times {V}_{1} \times {T}_{2}}{{P}_{2} \times {T}_{1}}$
We immediately see that we will get an answer in $\text{terms of volume}$, so we're batting on a firm wicket.
V_2=(6.0*atmxx20.0*mLxx273*K)/(2.0*atmxx546*K)=??*mL