# When the volume V_1 of a gas is halved at constant pressure, what is its new temperature if it began at 0^@ "C"?

Sep 28, 2016

I get $\text{300 K}$, or ${26.9}^{\circ} \text{C}$.

Any gas problem where you are not told what the gas is, or you are not told to use a real gas, can be assumed to be an ideal gas problem.

Therefore, you can use the ideal gas law:

$P V = n R T$

where:

• $P$ is the pressure . We can use $\text{atm}$.
• $V$ is the volume in $\text{L}$.
• $n$ is the $\boldsymbol{\text{mol}}$s of gas.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant for when you use pressure units of $\text{atm}$ (you do use a different one depending on your units of pressure and volume).
• $T$ is the temperature in $\text{K}$.

You are told that you are at constant pressure, so $\Delta P = {P}_{2} - {P}_{1} = 0$, meaning ${P}_{2} = {P}_{1}$. You are also given the initial temperature and volume.

Notice how if pressure is constant, you only have two variables changing: volume and temperature.

• Pressure is constant.
• The mols of gas are constant.
• The universal gas constant is... constant, no surprise.

You are then left with the following two equations:

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

When you divide these, you get:

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

So, to solve for the new temperature, you were told that the volume was halved. Therefore, ${V}_{2} = \frac{1}{2} {V}_{1}$. You then have:

$\frac{\frac{1}{2} \cancel{{V}_{1}}}{\cancel{{V}_{1}}} = {T}_{2} / {T}_{1}$

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

Note that ${0}^{\circ} \text{C" + 273.15 = "273.15 K}$, so we'll use $\text{600.15 K}$.

$\to \textcolor{b l u e}{{T}_{2}} = \frac{1}{2} \left(\text{600.15 K}\right)$

$=$ $\textcolor{b l u e}{\text{300.08 K}}$

Or, $300.08 - 273.15 \approx \textcolor{b l u e}{{26.9}^{\circ} \text{C}}$.