# The volume of a bicycle tire is 1.35 liters and the manufacturer recommends a tire pressure of 8.5 atm. If you want the bicycle tire to have the correct pressure at 20.0°C, what volume of air is required at STP?

Mar 5, 2016

$\text{11 L}$

#### Explanation:

The idea here is that you need to figure out what volume of gas held at STP conditions is needed in order for the tire to have a volume of $\text{1.35 L}$ at $\text{8.5 atm}$ and ${20.0}^{\circ} \text{C}$.

Since pressure, temperature, and volume change, you can use the combined gas law equation to find the volume of gas at STP.

The combined gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2}} |} \text{ }$, where

${P}_{1}$, ${V}_{1}$, ${T}_{1}$ - the pressure, volume, and absolute temperature of the gas at an initial state
${P}_{2}$, ${V}_{2}$, ${T}_{2}$ - the pressure, volume, and absolute temperature of a gas at a final state

So, STP conditions are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. To convert the pressure to atm and the temperature to Kelvin, use the conversion factors

$\text{1 atm " = " 101.325 kPa}$

$T \left[\text{K"] = t[""^@"C}\right] + 273.15$

You're starting with the gas under STP conditions, then changing its temperature to ${20.0}^{\circ} \text{C}$ and its pressure to $\text{8.5 atm}$.

Rearrange the combined gas law equation to solve for ${V}_{1}$

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2 \implies {V}_{1} = {P}_{2} / {P}_{1} \cdot {T}_{1} / {T}_{2} \cdot {V}_{2}$

Plug in your values to get

V_1 = (8.5 color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * ((273.15 + 0)color(red)(cancel(color(black)("K"))))/((273.15 + 20.0)color(red)(cancel(color(black)("K")))) * "1.35 L"

${V}_{1} = \text{10.834 L}$

Rounded to two sig figs, the number of sig figs you have for the pressure of the tire at ${20.0}^{\circ} \text{C}$, the answer will be

$V = \textcolor{g r e e n}{| \overline{\underline{\text{11 L}}} |}$

SIDE NOTE STP conditions are often given as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$, so if that is how STP conditions were defined ti you, simply redo the calculations using a pressure of $\text{1 atm}$ instead of a pressure of $\text{100 kPa}$.

Rounded to two sig figs, the answer will come out to be the same, $\text{11 L}$.