Question

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`?

Answer 1

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

`color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))" "`, 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 * T_1`

In your case, the volume of the gas must be kept constant at `x color(white)(a)"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^@"C" + 273.15 = "293.15 K"`

Now, the pressure is halved, which implies that

`P_2 = 1/2 * P_1`

Plug this into the equation to find

`T_2 = (1/2 * color(red)(cancel(color(black)(P_1))))/color(red)(cancel(color(black)(P_1))) * "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.