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?

Mar 8, 2016

$\text{352 K}$

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 $\text{205 kPa}$ to $\text{238 kPa}$. This tells you that the temperature of the must have increased as well.

Mathematically, you can write this as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

${P}_{1}$, ${T}_{1}$ - the pressure and temperature of the gas at an initial state
${P}_{2}$, ${T}_{2}$ - the pressure and temperature of the gas at final state

Rearrange the equation to solve for ${T}_{2}$

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \implies {T}_{2} = {P}_{2} / {P}_{1} \cdot {T}_{1}$

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

${T}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{352 K} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As predicted, the temperature of the air increased.