# The gas inside of a container exerts "84 Pa" of pressure and is at a temperature of "320 K". If the pressure in the container changes to "64 Pa" with no change in the container's volume, what is the new temperature of the gas?

Apr 1, 2016

One way to do this is to use the ideal gas law and derive an equation.

$P V = n R T$

where:

• $P$ is the pressure in $\text{Pa}$, let's say. It doesn't matter in this case.
• $V$ 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 $\text{K}$.

Then, we would say that if state $1$ represented an initial state and state $2$ a final state, we have ${P}_{1} \to {P}_{2}$ and ${T}_{1} \to {T}_{2}$, but ${V}_{1} = {V}_{2} = V$ and ${n}_{1} = {n}_{2} = n$.

With this information we get two ideal gas law relationships:

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

Therefore, to find the new temperature, we can divide these to get:

${P}_{1} / {P}_{2} = {T}_{1} / {T}_{2}$

$\setminus m a t h b f \left({T}_{2} = {T}_{1} \cdot {P}_{2} / {P}_{1}\right)$

So, now that we have the final equation, we can acquire ${T}_{2}$ for an ideal gas.

$\textcolor{b l u e}{{T}_{2}} = \text{320 K} \times \left(\frac{84}{64}\right)$

$=$ $\textcolor{b l u e}{\text{420 K}}$