Question #d9848

1 Answer
Aug 2, 2017

Answer:

#"mol" = "volume"/22.41#

Explanation:

The ideal gas law states

#ul(PV = nRT#

where

  • #P# is the pressure (in #"atm"#) of the gas

  • #V# is the volume (in #"L"#) the gas occupies

  • #n# is the quantity (in #"mol"#) of gas present

  • #R# is the universal gas constant, equal to #0.082057("L"·"atm")/("mol"·"K")#

  • #T# is the absolute temperature (in #"K"#) of the gas (absolute temperature indicates units of Kelvin)

Standard temperature and pressure (STP) conditions are commonly used in chemistry as

  • #ul(273.15color(white)(l)"K"#

  • #ul(1color(white)(l)"atm"#

(Standard pressure, since the year 1982, has been defined as #1# #"bar"# (#0.9869# #"atm"#), but a lot of instructors teach it as #1# #"atm"#. The difference is small, but can cause differing calculations, so be sure to know which standard pressure you are to be using.)

Plugging these and the constant #R# into the equation, we have

#(1color(white)(l)"atm")(V) = n(0.082057("L"·"atm")/("mol"·"K"))(273.15color(white)(l)"K")#

We're asked to use the gas law to find the moles (#n#) from a given volume (#V#), so let's eliminate the units in the above expression, and rearrange to solve for #n#:

#(1)(V) = (22.41)(n)#

#color(red)(ulbar(|stackrel(" ")(" "n = V/22.41" ")|)#

Does the number #22.41# (or #22.4#) ring a bell, perhaps? This is where it comes from!