# Question 624ba

Mar 17, 2016

$\text{3.00 moles}$

#### Explanation:

The idea here is that one mole of any ideal gas that is kept under Standard Temperature and Pressure, STP, conditions will occupy $\text{22.4 L } \to$ this is known as the molar volume of a gas at STP.

This can be derived using the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, 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 $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

STP conditions are defined as a pressure of $\text{1 atm}$ and a temperature of $\text{273.15 K}$. Plug these values into the ideal gas law equation and solve for $\frac{V}{n}$

$P V = n R T \implies \frac{V}{n} = \frac{R T}{P}$

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 $\text{22.4 L}$. So, if $\text{22.4 L}$ correspond to one mole of any ideal gas, it follows that $\text{67.2 L}$ will correspond to

$67.2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L"))) * overbrace("1 mole"/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)"3.00 moles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

SIDE NOTE It's worth noting that the current STP conditions are defined as a pressure of $\text{100 kPa}$ and a temperature of $\text{273.15 K}$.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.7 L}$.