# If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolute temperature of the gas be affected?

##### 1 Answer

The temperature of the gas will decrease by a factor of

#### Explanation:

The *pressure* and *temperature* of a gas have a **direct relationship** when pressure and number of moles of gas are **kept constant** **Charles' Law**.

Simply put, when the pressure of a given sample of gas is kept constant, **increasing** the volume of the gas will result in an **increase** in temperature

Similarly, **decreasing** the volume of the gas, as you have in your example, will result in a **decrease** in temperature.

Mind you, the *temperature* of the gas is actually its **absolute temperature**, i.e. its temperature expressed in *Kelvin*,

Now, the thing to remember here is that decreasing or increasing the volume of the gas **by a factor** **by the same factor**

Mathematically, this can be shown by using the equation for Charles' Law

#color(blue)(|bar(ul(color(white)(a/a)V_1/T_1 = V_2/T_2color(white)(a/a)|)))" "# , where

Let's say that your starting volume is

#V_2 = 1/3 * V_1 -># the volumedecreasesby a factor of#3#

Rearrange the equation to solve for

#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#

You will thus have

#T_2 = (1/3 * color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * T_1 = color(green)(|bar(ul(color(white)(a/a)color(black)(1/3 * T_1)color(white)(a/a)|)))#

As you can see, **decreasing** the volume by a factor of **absolute temperature** to **decrease** by a factor of