# If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change?

##### 1 Answer

#### Explanation:

Even without doing any calculations, you should be able to look at the values given to you and predict that the volume of the gas will **decrease** as temperature *decreases*.

When pressure and number of moles of gas are **held constant**, the volume of a gas and its temperature have a **direct relationship** - this is known as **Charles' Law**.

As you know, gas pressure is caused by the collisions that take place between the molecules of gas and the walls of the container.

The **more powerful and frequent** these collisions are, the **higher** the pressure of the gas.

Now, temperature is a measure of the *average kinetic energy* of the gas molecules. When you **decrease** temperature, you're essentially decreasing the average speed with which these molecules hit the walls of the container.

Because molecules are hitting the walls of the container with *less force*, you need these collisions to be **more frequent** in order for pressure to be **constant**.

This means that the volume of the gas must **decrease** as well, since the same number of molecules in a *smaller volume* will result in more frequent collisions with the walls of the container.

So, when temperature *decreases*, volume **decreases** as well.

Mathematically, this is written as

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

Now, it's **very important** to remember that you must use **absolute temperature**, i.e. the temperature expressed in *Kelvin*.

To go from *degrees Celsius* to *Kelvin*, use the conversion factor

#color(blue)(|bar(ul(color(white)(a/a)T["K"] = t[""^@"C"] + 273.15color(white)(a/a)|)))#

So, rearrange the equation for Charles' Law and solve for

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

Plug in your values to get

#V_2 = ((273.15 + 25)color(red)(cancel(color(black)("K"))))/((273.15 + 325)color(red)(cancel(color(black)("K")))) * "6.80 L" = "3.3895 L"#

You need to round this off to two **sig figs**, the number of sig figs you have for the final temperature of the gas

#V_2 = color(green)(|bar(ul(color(white)(a/a)"3.4 L"color(white)(a/a)|)))#