# Question #675cd

Jan 4, 2015

I assume the question refers to 4 L of gas, and asks for the new volume after the pressure and the temperature are doubled.

This can be solved by using the combined gas law,

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

You start with a pressure of ${P}_{1}$ and end with a pressure of ${P}_{2} = 2 \cdot {P}_{1}$. Likewise, the initial temperature is ${T}_{1}$, and the final temperature will be ${T}_{2} = 2 \cdot {T}_{1}$.

So, we can determine ${V}_{2}$ from the combined gas law equation

${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1} \implies {V}_{2} = {P}_{1} / \left(2 \cdot {P}_{1}\right) \cdot \frac{2 \cdot {T}_{1}}{T} _ 1 \cdot {V}_{1}$

It's evident that ${V}_{2} = {V}_{1}$, since both the pressure and the temperature terms cancel out.

This would have also been the case if the question said 4 moles of gas, since the combined gas law assumes that the number of moles is constant.

You would get the same result, ${V}_{\text{final") = V_("initial}}$.

SInce the question provides a value for ${V}_{1}$, the answer is

${V}_{2} = {V}_{1} = 4$ $\text{L}$.