Question

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?

Answer 1

`"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 `"1.35 L"` at `"8.5 atm"` and `20.0^@"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

`color(blue)(|bar(ul((P_1V_1)/T_1 = (P_2V_2)/T_2))|)" "`, 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 `"100 kPa"` and a temperature of `0^@"C"`. To convert the pressure to atm and the temperature to Kelvin, use the conversion factors

`"1 atm " = " 101.325 kPa"`

`T["K"] = t[""^@"C"] + 273.15`

You're starting with the gas under STP conditions, then changing its temperature to `20.0^@"C"` and its pressure to `"8.5 atm"`.

Rearrange the combined gas law equation to solve for `V_1`

`(P_1V_1)/T_1 = (P_2V_2)/T_2 implies V_1 = P_2/P_1 * T_1/T_2 * 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 = "10.834 L"`

Rounded to two sig figs, the number of sig figs you have for the pressure of the tire at `20.0^@"C"`, the answer will be

`V = color(green)(|bar(ul("11 L"))|)`

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

Rounded to two sig figs, the answer will come out to be the same, `"11 L"`.