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

\(7.4\% \)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 \({T_1} = {15^ \circ }{\rm{C}} = 288\;{\rm{K}}\).

The final temperature is \({T_2} = {38^ \circ }{\rm{C}} = 311\;{\rm{K}}\).

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.

\(\begin{array}{c}nT = {\rm{constant}}\\{n_1}{T_1} = {n_2}{T_2}\\\frac{{{n_1}}}{{{n_2}}} = \frac{{{T_2}}}{{{T_1}}}\\\left( {\frac{{{n_1} - {n_2}}}{{{n_1}}}} \right) \times 100\% = \left( {\frac{{{T_2} - {T_1}}}{{{T_2}}}} \right) \times 100\% \end{array}\)

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

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

Hence, \(7.4\% \)gas must be removed.