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The air in a bicycle tire is bubbled through water and collected at 25°C. If we assume that the air that has been collected at 25°C has a total volume of 5.45 L and pressure of 745 torr, calculate the moles of air that were stored in the bicycle tire?

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 quantity 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:

\[n_{air}=0.3545mol\]

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