Gas Laws Fact Sheet
\begin{tabular}{|l|l|}
\hline Ideal gas law & [tex]$P V = n R T$[/tex] \\
\hline Ideal gas constant & \begin{tabular}{l}
[tex]$R = 8.314$[/tex] \\
or \\
[tex]$R = 0.0821$[/tex]
\end{tabular} \\
\hline Standard atmospheric pressure & [tex]$1 \, \text{atm} = 101.3 \, \text{kPa}$[/tex] \\
\hline Celsius to Kelvin conversion & [tex]$K = {}^{\circ}C + 273.15$[/tex] \\
\hline
\end{tabular}

Select the correct answer.

A volleyball is full of pressurized air. The air temperature is [tex]$24.6^{\circ}C$[/tex]. The volleyball's absolute pressure is [tex]$130.75 \, \text{kPa}$[/tex], and its volume is [tex]$5.27 \, \text{liters}$[/tex]. How many moles of air are inside the volleyball?

A. [tex]$0.278 \, \text{mol}$[/tex]
B. [tex]$3.37 \, \text{mol}$[/tex]
C. [tex]$3.59 \, \text{mol}$[/tex]
D. [tex]$25.2 \, \text{mol}$[/tex]
E. [tex]$28.0 \, \text{mol}$[/tex]



Answer :

To determine the number of moles of air inside the volleyball, we will use the Ideal Gas Law, represented by the equation:

[tex]\[ PV = nRT \][/tex]

where:
- [tex]\( P \)[/tex] is the pressure in kilopascals (kPa),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the ideal gas constant in units of [tex]\( \text{L·kPa/(mol·K)} \)[/tex],
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).

Given data:
- Temperature in Celsius: [tex]\(24.6^\circ \text{C}\)[/tex]
- Pressure in kilopascals: [tex]\(130.75 \text{ kPa}\)[/tex]
- Volume in liters: [tex]\(5.27 \text{ L}\)[/tex]
- Ideal gas constant: [tex]\(R = 8.314 \text{ L·kPa/(mol·K)}\)[/tex]

First, we need to convert the temperature from Celsius to Kelvin:
[tex]\[ T = 24.6 + 273.15 = 297.75 \text{ K} \][/tex]

Next, we rearrange the ideal gas law equation to solve for [tex]\( n \)[/tex] (number of moles):
[tex]\[ n = \frac{PV}{RT} \][/tex]

Substituting the given values into the equation:
[tex]\[ n = \frac{(130.75 \text{ kPa}) \times (5.27 \text{ L})}{(8.314 \text{ L·kPa/(mol·K)}) \times (297.75 \text{ K})} \][/tex]

From the calculation, we obtain:
[tex]\[ n \approx 0.2783495492918886 \text{ mol} \][/tex]

Rounding this to three significant figures, we get:
[tex]\[ n \approx 0.278 \text{ mol} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{0.278 \text{ mol}} \][/tex]

So, the correct option is:
A. 0.278 mol