# 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#?

##### 1 Answer

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

**absolute temperature** of the gas at an initial state

**absolute temperature** of the gas at final state

Rearrange the equation to solve for

#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 **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*.