Question f6a82

Jun 9, 2017

Here's what I got.

Explanation:

The idea here is that STP conditions are usually given to you as a pressure of $\text{1 atm}$ and a temperature of

${0}^{\circ} \text{C" = 0^@"C" + 273.15 = "273.15 K}$

Under these specific conditions for pressure and temperature, $1$ mole of any ideal gas occupies $\text{22.4 L}$. This is referred to as the molar volume of a gas at STP.

Now, room conditions are usually given to you as a pressure of $\text{1 atm}$ and a temperature of

${20}^{\circ} \text{C" + 273.15 = "293.15 K}$

our goal here is to figure out the volume occupied by $1$ mole of any ideal gas at room temperature. To do that, use the ideal gas law equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange the equation

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

Plug in your values to get

V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("atm"))))

$\frac{V}{n} = {\text{24.07 L mol}}^{- 1}$

This means that $1$ mole of any ideal gas occupies $\text{24.07 L}$ under room conditions.

Now all you have to do is use the molar mass of nitrogen gas to determine how many moles you have in your sample

35 color(red)(cancel(color(black)("g"))) * "1 mole n"_2/(28.0134color(red)(cancel(color(black)("g")))) = "1.249 moles N"_2#

and use the molar volume of a gas at room conditions to find the volume it occupies

$1.249 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles N"_2))) * "24.07 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(darkgreen)(ul(color(black)("30. L}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of nitrogen as.

SIDE NOTE Plug in the values for pressure and temperature you have at STP into this equation

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

You should end up with

$\frac{V}{n} = {\text{22.4256 mol L"^(-1) ~~ "22.4 mol L}}^{- 1}$

This is why we can say that 1 mole of any ideal gas occupies $\text{22.4 L}$ under STP conditions.