The aim of this question is to find the quantity of air in moles that were stored in a bicycle tire.

To calculate the quantity of gas stored at a certain pressure and temperature, we assume the given gas is an ideal gas, and we will use the concept of **Ideal Gas Law.**

An **Ideal Gas** is a gas comprising particles that neither attract nor repel each other and take up no space (have no volume). They move independently and interact with each other in the form of elastic collisions only.

**Ideal Gas Law** or **General Gas Equation** is the equation of the state of an ideal gas determined by the parameters like **Volume**, **Pressure**, and **Temperature**. It is written as shown below:

\[PV=nRT\]

Where:

$P$ is the given **pressure** of the ideal gas.

$V$ is the given **volume** of the ideal gas.

$n$ is the **quantit****y** of ideal gas in **moles**.

$R$ is the **gas constant**.

$T$ is the **temperature **in** Kelvin** $K$.

## Expert Answer

Given as:

The **pressure of air** after passing through water $P_{gas}=745\ torr$

**Temperature** $T=25^{\circ}C$

**Volume** $V=5.45$ $L$

We need to find the **number of moles of air** $n_{air}$

We also know that:

**Vapor pressure of water** $P_w$ at $25^{\circ}C$ is $0.0313atm$, or $23.8$ $mm$ $of$ $Hg$

**Gas constant** $R=\dfrac{0.082atmL}{Kmol}$

In the first step, we will convert the given values into **SI units.**

$(a)$ **Temperature** must be in **Kelvin** $K$

\[K=°C+273.15\]

\[K=25+273.15=298.15K\]

$(b)$ **Pressure** $P_{gas}$ must be in **atmosphere** $atm$

\[760\ torr=1\ atm\]

\[P_{gas}=745\ torr=\frac{1\ atm}{760}\times745=0.9803atm\]

In the second step, we will use the **Dalton’s Law of Partial Pressure** to calculate the pressure of air.

\[P_{gas}=P_{air}+P_w\]

\[P_{air}=P_{gas}-P_w\]

\[P_{air}=0.9803atm-0.0313atm=0.949atm\]

Now, by utilizing the **Idea Gas Law**, we will calculate the **number of moles of air** $n_{air}:$

\[P_{air}V=n_{air}RT\]

\[n_{air}=\frac{P_{air}V}{RT}\]

By substituting the given and calculated values:

\[n_{air}=\frac{0.949\ atm\times5.45L}{(\dfrac{0.082\ atmL}{Kmol})\times298.15K}\]

By solving the equation and canceling the units, we get:

\[n_{air}=0.2115mol\]

## Numerical Results

The **number of moles of air** that were stored in the bicycle is $n_{air}=0.2115mol$.

## Example

**Air stored in a tank** is **bubbled** through a water beaker and collected at **$30^{\circ}C$** having a volume of **$6L$** at a pressure of **$1.5atm$.** Calculate the **moles of air** that were stored in the tank.

Given As:

The **pressure of air** after passing through water $P_{gas}=1.5\ atm$

**Temperature** $T=30^{\circ}C=303.15K$

**Volume** $V=6$ $L$

We need to find the **number of moles of air** $n_{air}$ stored in the tank.

We also know that:

**Vapor pressure of water** $P_w$ at $25^{\circ}C$ is $0.0313atm$, or $23.8$ $mm$ $of$ $Hg$

**Gas constant** $R=\dfrac{0.082atmL}{Kmol}$

\[P_{gas}=P_{air}+P_w\]

\[P_{air}=P_{gas}-P_w\]

\[P_{air}=1.5atm-0.0313atm=1.4687atm\]

Now, by utilizing the **Idea Gas Law**, we will calculate the **number of moles of air** $n_{air}:$

\[P_{air}V=n_{air}RT\]

\[n_{air}=\frac{P_{air}V}{RT}\]

By substituting the given and calculated values:

\[n_{air}=\frac{1.4687\ atm\times6L}{(\dfrac{0.082\ atmL}{Kmol})\times303.15K}\]

By solving the equation and canceling the units, we get:

\[n_{air}=0.3545mol\]