A soccer ball of constant volume \(2.24 \mathrm{~L}\) is pumped up with air to a gauge pressure of \(13 \mathrm{lb} /\) in. \({ }^{2}\) at \(20.0^{\circ} \mathrm{C}\). The molar mass of air is about \(29 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of air are in the ball? (b) What mass of air is in the ball? (c) During the game, the temperature rises to \(30.0^{\circ} \mathrm{C}\). What mass of air must be allowed to escape to bring the gauge pressure back to its original value?

When dealing with gases, it's essential to use the correct pressure values. Gauge pressure does not include atmospheric pressure. Absolute pressure, however, does. In this exercise, the gauge pressure of the soccer ball is given as 13 lb/in². To get absolute pressure, add atmospheric pressure (approx 14.7 lb/in²).

This gives:

\[ P_{abs} = 13 \text{\text{lb/in}}^2 + 14.7 \text{\text{lb/in}}^2 = 27.7 \text{\text{lb/in}}^2 \]
For calculations involving the Ideal Gas Law, convert pressure into Pascals (Pa). Using the conversion factor 1 lb/in² = 6894.76 Pa:
\[ P_{abs} = 27.7 \text{\text{lb/in}}^2 \times 6894.76 \text{Pa/(\text{lb/in}}^2) = 190,577 \text{\text{Pa}} \]
Understanding this conversion is vital. It ensures consistency, especially when dealing with the Ideal Gas Law.

Calculating the number of moles () in a gas using the Ideal Gas Law simplifies understanding chemical and physical behaviors. The Ideal Gas Law is represented as:
\[ PV = nRT \]
Here, P is pressure (Pa), V is volume (m³), n is number of moles, R is the gas constant (8.314 J/(mol·K)), and T is temperature (K).

In this problem, convert the given volume of the soccer ball from liters to cubic meters:
\[ V = 2.24 \text{L} = 2.24 \times 10^{-3} \text{m}^3 \] Convert the temperature from Celsius to Kelvin:
\[ T = 20.0^\text{ C} + 273 = 293 \text{K} \]
Then solve for the number of moles by rearranging the Ideal Gas Law formula:
\[ n = \frac{PV}{RT} \]
Using the values:
\[ n = \frac{190,577 \text{Pa} \times 2.24 \times 10^{-3} \text{m}^3}{8.314 \text{J/(mol·K)} \times 293 \text{K}} \]
This computes to:
\[ n = 0.186 \text{moles} \]
This fundamental process is crucial for further calculations in gas behavior.

The Ideal Gas Law is one of the cornerstone gas laws. It describes how gases behave under different conditions of pressure, volume, and temperature. When temperature changes, it affects the amounts of gas (in moles) required to maintain the same pressure.

For part of this exercise, the temperature rises to 30.0°C (303 K).
To determine the new number of moles needed to keep the same pressure, use the ratio of initial and final temperatures:
\[ n_2 = n_1 \times \frac{T_2}{T_1} \]
With:
\[ n_1 = 0.186 \text{moles}, \text{T}_1 = 293 \text{K}, \text{T}_2 = 303 \text{K} \] Compute:
\[ n_2 = 0.186 \text{moles} \times \frac{303}{293} = 0.192 \text{moles} \]
The difference in moles indicates how much air must be released to maintain the same pressure:
\[ n_{diff} = n_2 - n_1 = 0.192 \text{moles} - 0.186 \text{moles} = 0.006 \text{moles} \]
Convert this to mass using the molar mass of air (29 g/mol):
\[ \text{Mass}_{diff} = n_{diff} \times \text{Molar Mass} = 0.006 \text{moles} \times 29 \text{g/mol} = 0.174 \text{g} \]
This helps maintain consistent pressure by releasing 0.174g of air when temperature increases.