# How do you calculate the temperature of the gas at this new pressure?

## A sample of an ideal gas is sealed in a container whose volume cannot change. Initially it has a pressure of 2.5 x 10^5 Pa and its temperature -20 oC.The gas is cooled until the pressure is reduced to 1.0 x 10^5 Pa.

Jun 20, 2016

Since you are talking about volume, pressure, and temperature, we can naturally use the ideal gas law.

$\setminus m a t h b f \left(P V = n R T\right)$

where:

• $P$ is the pressure in $\text{Pa}$.
• $V$ is the volume in $\text{L}$.
• $n$ is $\text{mol}$s of gas, which has not changed throughout the process.
• $R$ is the universal gas constant. We won't need to use this because it is a constant throughout the process.
• $T$ is the temperature in "K".

You are told that the volume cannot change, so the volume stays constant. Note your variables:

${P}_{1} = 2.5 \times {10}^{5} \text{Pa}$
${P}_{2} = 1.0 \times {10}^{5} \text{Pa}$
${T}_{1} = - {20}^{\circ} \text{C" = "253.15 K}$
T_2 = ?#

So, set up your equations using these variables. We know that $V$, $n$, and $R$ are the same across both equations because the container is closed (constant $n$) and rigid (constant $V$).

(Obviously, the universal gas constant is constant.)

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

Therefore, we can divide these equations to solve for ${T}_{2}$.

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

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

$\implies \textcolor{b l u e}{{T}_{2}} = {T}_{1} \times \frac{{P}_{2}}{{P}_{1}}$

$= \left(\text{253.15 K")xx(1.0xx10^5 cancel"Pa")/(2.5xx10^5 cancel"Pa}\right)$

$=$ $\textcolor{b l u e}{\text{101.26 K}}$

$=$ $\textcolor{b l u e}{- {171.89}^{\circ} \text{C}}$