# What happens to the volume of an ideal gas if the pressure halves and temperature doubles?

Jul 27, 2017

Consider the ideal gas law...

$\boldsymbol{P V = n R T}$

where $P$, $V$, $n$, $R$, and $T$ are pressure in $\text{atm}$, volume in $\text{L}$, mols in $\text{mol}$s, the universal gas constant in $\text{L"cdot"atm/mol"cdot"K}$, and temperature in $\text{K}$, respectively.

You should know how to manipulate this equation in general for an exam---this is fair play.

METHOD 1

Volume is a state function, and hence one can break down two changes in related variables into two separate steps.

Since the pressure halved and the temperature doubled, consider their stacked effects in two steps...

• Step 1: halving the pressure at constant temperature means half the volume compression, so the volume doubles after step 1.
• Step 2: doubling the temperature at constant pressure doubles the volume after step 2.

METHOD 2

Let $P \to \frac{1}{2} P$ and $T \to 2 T$. Then $V \to c V$, and we must find that constant $c$ to see how $V$ changed to maintain the equality.

$\frac{1}{2} P \left(c V\right) = n R \left(2 T\right)$

$\textcolor{b l u e}{\overline{\underline{\stackrel{\text{ ")(|" "cPV = 4nRT" }}{|}}}}$

Thus, since $P V = n R T$, $c P V = c N R T$, and $c = 4$. As a result, we see that the volume quadrupled.

METHOD 3

Another way to do this is to set initial and final states.

${P}_{1} {V}_{1} = n R {T}_{1}$

${P}_{2} {V}_{2} = n R {T}_{2}$

And we say, what is ${V}_{2} / {V}_{1}$ if ${P}_{2} = \frac{1}{2} {P}_{1}$ and ${T}_{2} = 2 {T}_{1}$? Plug in these substitutions. Solve for ${V}_{2}$ and then solve for ${V}_{1}$.

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

$\implies \textcolor{b l u e}{\overline{\underline{\stackrel{\text{ ")(|" "V_2 = (color(red)(4)nRT_1)/P_1" }}{|}}}}$

For ${V}_{1}$, we obtain:

$\implies \textcolor{b l u e}{\overline{\underline{\stackrel{\text{ ")(|" "V_1 = (nRT_1)/P_1" }}{|}}}}$

By inspection, we see that ${V}_{2} = 4 {V}_{1}$, so the volume has quadrupled.