# A gas sample has a volume of x "dm"^3 at 20^@"C". If the pressure is halved, what temperature is required to maintain the volume at x "dm"^3?

Jul 6, 2016

Here's what I got.

#### Explanation:

The idea here is that when the number of moles and the volume of the gas are kept constant, pressure and temperature have a direct relationship as described by Gay Lussac's Law.

In other words, increasing the pressure by a factor will cause the volume to increase by the same factor. Likewise, decreasing the pressure by a factor will cause the volume to decrease by the same factor. Mathematically, this is written as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

${P}_{1}$, ${T}_{1}$ - the pressure and absolute temperature of the gas at an initial state
${P}_{2}$, ${T}_{2}$ - the pressure and absolute temperature of the gas at final state

Rearrange the equation to solve for ${T}_{2}$

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \implies {T}_{2} = {P}_{2} / {P}_{1} \cdot {T}_{1}$

In your case, the volume of the gas must be kept constant at $x \textcolor{w h i t e}{a} {\text{dm}}^{3}$. The temperature of the gas must be expressed in Kelvin, so make sure that you convert it before doing anything else

${T}_{1} = {20}^{\circ} \text{C" + 273.15 = "293.15 K}$

Now, the pressure is halved, which implies that

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

Plug this into the equation to find

${T}_{2} = \frac{\frac{1}{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}} \cdot \text{293.15 K" = "146.6 K}$

Convert this back to degrees Celsius

t_2 = "146.6 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)color(black)(-130^@"C")color(white)(a/a)|)))

I'll leave the answer rounded to two sig figs.

So, when the pressure of a gas is halved, the only way to keep its volume constant is to halve its absolute temperature.