Question

A sample of propane, a component of LP gas, has a volume of 35.3 L at 315 K and 922 torr. What is its volume at STP?

Answer 1

`"37.6 L"`

Explanation:

STP conditions are defined as a pressure of

`"100 kPa" = 100 color(red)(cancel(color(black)("kPa"))) * (1color(red)(cancel(color(black)("atm"))))/(101.325color(red)(cancel(color(black)("kPa")))) * "760 torr"/(1color(red)(cancel(color(black)("atm")))) ="750.1 torr"`

and a temperature of

`0^@"C" = 0^@"C" + 273.15 = "273.15 K"`

so, right from the start, you can say that both the temperature and the pressure of the gas are changing.

This means that your tool of choice here will be the combined gas law equation

`color(blue)(ul(color(black)((P_1V_1)/T_1 = (P_2V_2)/T_2)))`

Here

• `P_1`, `V_1`, `T_1` are the pressure, volume, and absolute temperature of the gas at an initial state
• `P_2`, `V_2`, `T_2` are the pressure, volume, and absolute temperature of the gas at a final state

Your goal is to find the volume of the gas at STP, so rearrange the equation to solve for `V_2`

`(P_1V_1)/T_1 = (P_2V_2)/T_2 implies V_2 = P_1/P_2 * T_2/T_1 * V_1`

Plug in your values to get

`V_2 = (922 color(red)(cancel(color(black)("torr"))))/(750.1color(red)(cancel(color(black)("torr")))) * (273.15color(red)(cancel(color(black)("K"))))/(315color(red)(cancel(color(black)("K")))) * "35.3 L"`

`V_2 = color(darkgreen)(ul(color(black)("37.6 L")))`

The answer is rounded to three sig figs.