Question

Question #624ba

Answer 1

`"3.00 moles"`

Explanation:

The idea here is that one mole of any ideal gas that is kept under Standard Temperature and Pressure, STP, conditions will occupy `"22.4 L " ->` this is known as the molar volume of a gas at STP.

This can be derived using the ideal gas law equation

`color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "`, where

`P` - the pressure of the gas
`V` - the volume occupied by the gas
`n` - the number of moles of gas present
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

STP conditions are defined as a pressure of `"1 atm"` and a temperature of `"273.15 K"`. Plug these values into the ideal gas law equation and solve for `V/n`

`PV = nRT implies V/n = (RT)/P`

`V/n = (0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 273.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm")))) = "22.4 L mol"^(-1)`

This tells you that every mole of an idea lgas that is being kept under STP conditions will occupy `"22.4 L"`.

So, if `"22.4 L"` correspond to one mole of any ideal gas, it follows that `"67.2 L"` will correspond to

`67.2color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)"3.00 moles"color(white)(a/a)|)))`

SIDE NOTE It's worth noting that the current STP conditions are defined as a pressure of `"100 kPa"` and a temperature of `"273.15 K"`.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies `"22.7 L"`.