# 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?

##### 1 Answer

#### Answer:

#### 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

#(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**.