Question

(II) A tire is filled with air at 15°C to a gauge pressure of 230 kPa. If the tire reaches a temperature of 38°C, what fraction of the original air must be removed if the original pressure of 230 kPa is to be maintained?

Short answer

$7.4\mathrm{\%}$gas must be removed.

Step-by-step solution

Concepts

The ideal gas law is $PV=nRT$ for n mole of ideal gas.

For this problem, you have to use the equation$n_1{T}_1=n_2{T}_2$.

Given data 

The initial temperature is $T_1={15}^{\circ }\mathrm{C}=288\mathrm{K}$.

The final temperature is $T_2={38}^{\circ }\mathrm{C}=311\mathrm{K}$.

Let $T_2$ be the temperature of the gas when the pressure gets doubled.

Assume that there are $n_1$ moles of gas initially, but there are $n_2$ moles of the gas inside the tire at the final temperature.

Assume that the volume of the tire does not change.

Calculation

Rearranging the ideal gas equation, you get,

$nT=\frac{PV}{R}$.

For this problem, the volume and the pressure of the gas were kept constant.

$\begin{array}{c}nT=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\\ n_1{T}_1=n_2{T}_2\\ \frac{n_1}{n_2}=\frac{T_2}{T_1}\\ \left(\frac{n_1-n_2}{n_1}\right)\times 100\mathrm{\%}=\left(\frac{T_2-T_1}{T_2}\right)\times 100\mathrm{\%}\end{array}$

Now, substituting the values into the above equation, you get the following:

$\begin{array}{c}\left(\frac{n_1-n_2}{n_1}\right)\times 100\mathrm{\%}=\left(\frac{T_2-T_1}{T_2}\right)\times 100\mathrm{\%}\\ =\left(\frac{311\mathrm{K}-288\mathrm{K}}{311\mathrm{K}}\right)\times 100\mathrm{\%}\\ =7.4\mathrm{\%}\end{array}$

Hence, $7.4\mathrm{\%}$gas must be removed.