# At a pressure of "185 mmHg" and a temperature of 55^@"C", what volume would 2.55xx10^28 molecules of nitrogen gas occupy?

Dec 3, 2014

You will need to use the ideal gas law in order to answer this question. The ideal gas law is represented by the following equation::

$\text{P""V}$ = $\text{nRT}$, where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature in Kelvins.

First there are two conversions that must be made. The temperature in degrees Celsius needs to be converted to Kelvins. This is done by adding 273 to the Celsius temperature:
$\text{55.0"^o"C" + 273 = "328K}$

The number of molecules of nitrogen gas must be converted to moles of nitrogen gas. ${\text{1 mol N"_2 "molecules" = "6.022 x 10"^23 "molecules N}}_{2}$.

$\text{2.55 x 10"^28 "molecules N"_2}$ x $\text{1mol/6.022 x 10"^23}$ = ${\text{42345 mol N}}_{2}$

Now we're ready to solve the problem.

Given/Known:
$\text{P = 185 mmHg}$
${\text{n = 42345 mol N}}_{2}$
$\text{R}$ = ${\text{62.36367 L mmHg K"^(-1) "mol}}^{- 1}$
$\text{T}$ = $\text{328K}$

Unknown:
$\text{V}$

Equation:
$\text{P""V}$ = $\text{nRT}$

Solution : Rearrange the equation to isolate $\text{V}$ by dividing both sides by $\text{P}$.

$\text{V}$ = $\text{nRT"/"P}$ = $\text{(42345mol)(62.36367L mmHg K"^(-1)"mol"^(-1)}$)$\text{(328K)}$$\div$$\text{185 mmHg}$

$\text{V}$ = $\text{4.68 x 10"^6 "L}$ (rounded to three significant figures)