# The pressure in a car tire is 205 kPa at 303 K. After a long drive, the pressure is 238 kPa. What is the temperature of the air in the tire?

##### 1 Answer

#### Answer:

#### Explanation:

You can assume that the volume of the tire and the amount of air it contains **remain constant**, which means that you can focus solely on the relationship that exists between *pressure* and *temperature* under these conditions.

More specifically, you should know that when volume and number of moles are **kept constant**, temperature and pressure have a **direct relationship** - this is known as **Gay Lussac's Law**.

So, when temperature **increases**, pressure **increases** as well. Likewise, when temperature **decreases**, pressure **decreases** as well.

In your case, the pressure in the tire increased from **must have increased** as well.

Mathematically, you can write this as

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

Rearrange the equation to solve for

#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#

Plug in your values to get

#T_2 = (238color(red)(cancel(color(black)("kPa"))))/(205color(red)(cancel(color(black)("kPa")))) * "303 K" = "351.78 K"#

Rounded to three **sig figs**, the answer will be

#T_2 = color(green)(|bar(ul(color(white)(a/a)"352 K"color(white)(a/a)|)))#

As predicted, the temperature of the air *increased*.