The volume of a gas is 93 mL when the temperature is 91°C. If the temperature is reduced to 0°C without changing the pressure, what is the new volume of the gas?

Mar 31, 2016

If we assume ideality, then we can use the ideal gas law:

$P V = n R T$

where:

• $P$ is the pressure in $\text{bar}$, let's say.
• $V$ is the volume in $\text{L}$.
• $n$ is the number of $\setminus m a t h b f \left(\text{mol}\right)$s of gas
• $R$ is the universal gas constant, which will be, based on our units, $\text{0.083145 L"cdot"bar/mol"cdot"K}$.
• $T$ is the temperature in units of $\text{K}$.

You can simply remember this equation instead of trying to remember all the smaller ones (Boyle's, Charles' and Gay-Lussac's laws, and Avogadro's Principle), and derive what you need.

We are looking at a change in volume due to a change in temperature, as stated in the question, and we assume that the pressure did NOT change.

Since $n$ and $R$ cannot change, that covers all of the variables that might or might not change, so we have closure knowing that we've accounted for all the variables.

Then, suppose we solve for $V$. If we then assign ${V}_{1}$ to the initial volume and ${V}_{2}$ to the final volume, we get the corresponding ${T}_{1}$ and ${T}_{2}$, but since ${P}_{1} = {P}_{2}$, we can still call it $P$.

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

Now, if we want to find ${V}_{2}$, we can divide these and solve for ${V}_{2}$.

${V}_{2} / {V}_{1} = \frac{\frac{n R {T}_{2}}{P}}{\frac{n R {T}_{1}}{P}} = {T}_{2} / {T}_{1}$

Therefore:

color(blue)(V_2 = V_1 T_2/T_1

So, you can use this formula to solve for ${V}_{2}$. Therefore, you should get:

$\textcolor{b l u e}{{V}_{2}}$

= "0.093 L" xx ("0 + 273.15 K"/"91 + 273.15 K")

$\approx$ $\textcolor{b l u e}{\text{0.070 L}}$