At a pressure of $185 mmHg$ $"185 mmHg"$ and a temperature of $55^{\circ} C$ $55^@"C"$ , what volume would $2.55 \times 10^{28}$ $2.55xx10^28$ molecules of nitrogen gas occupy?

Question

At a pressure of $185 mmHg$ $"185 mmHg"$ and a temperature of $55^{\circ} C$ $55^@"C"$ , what volume would $2.55 \times 10^{28}$ $2.55xx10^28$ molecules of nitrogen gas occupy?

1 Answer

Impact of this question

7525 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-12-03T05:51:39

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::

$P V$ $"P""V"$ = $nRT$ $"nRT"$ , 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:
${55.0}^{o} C + 273 = 328K$ $"55.0"^o"C" + 273 = "328K"$

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

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

Now we're ready to solve the problem.

Given/Known:
$P = 185 mmHg$ $"P = 185 mmHg"$
${n = 42345 mol N}_{2}$ $"n = 42345 mol N"_2$
$R$ $"R"$ = ${62.36367 L mmHg K}^{- 1} {mol}^{- 1}$ $"62.36367 L mmHg K"^(-1) "mol"^(-1)$
$T$ $"T"$ = $328K$ $"328K"$

Unknown:
$V$ $"V"$

Equation:
$P V$ $"P""V"$ = $nRT$ $"nRT"$

Solution : Rearrange the equation to isolate $V$ $"V"$ by dividing both sides by $P$ $"P"$ .

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

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

Science

Math

Humanities

... and beyond

At a pressure of $185 mmHg$ $"185 mmHg"$ and a temperature of $55^{\circ} C$ $55^@"C"$ , what volume would $2.55 \times 10^{28}$ $2.55xx10^28$ molecules of nitrogen gas occupy?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

At a pressure of 185 mmHg"185 mmHg" and a temperature of 55∘C55^@"C", what volume would 2.55×10282.55xx10^28 molecules of nitrogen gas occupy?

1 Answer

Related questions

Impact of this question

At a pressure of $185 mmHg$ $"185 mmHg"$ and a temperature of $55^{\circ} C$ $55^@"C"$ , what volume would $2.55 \times 10^{28}$ $2.55xx10^28$ molecules of nitrogen gas occupy?