Question

How many molecules of n2 are in a 300.0 ml container at 780 mmhg and 135°c?

Answer 1

To solve for the number of molecules, we need to first find the number of moles. Assuming ideal gas, we use the formula:

n = PV / RT

where P is pressure = 780 mmHg, V is volume = 300 mL = 0.3 L, T is temperature = 135°C = 408.15 K, R is gas constant = 62.36367 L mmHg / mol K

n = (780 mmHg) (0.3 L) / (62.36367 L mmHg / mol K) (408.15 K)

n = 9.19 x 10^-3 mol

 

Using Avogadros number, we calculate the number of molecules.

molecules = 9.19 x 10^-3 mol * 6.022 x 10^23 molecules / mol

molecules = 5.54 x 10^21 molecules

Answer 2

$\boxed{{5{.4 \times 1}}{{\text{0}}^{{\text{21}}}}\;{\text{molecules}}}$ of nitrogen are present in a 300 mL container at 780 mm Hg and $135\;^\circ{\text{C}}$.

Further Explanation:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is just a theoretical concept and practically no such gas exists. But gases tend to behave almost ideally at a higher temperature and lower pressure.

Ideal gas law is considered as the equation of state for any hypothetical gas. Here, we assume nitrogen to be an ideal gas. So the expression for the ideal gas equation of nitrogen is as follows:

${\text{PV}} = {\text{nRT}}$                                    ......(1)

Here, P is the pressure of nitrogen.

V is the volume of nitrogen.

T is the absolute temperature of nitrogen.

n is the number of moles of nitrogen.

R is the universal gas constant.

Rearrange equation (1) to calculate the number of moles of nitrogen.

${\text{n}} = \frac{{{\text{PV}}}}{{{\text{RT}}}}$        ......(2)

Firstly, the temperature is to be converted into K. The conversion factor for this is,

${\text{0 }}^\circ {\text{C}} = {\text{273 K}}$

So the temperature of nitrogen is calculated as follows:

$\begin{aligned}{\text{Temperature}}\left( {\text{K}}\right)&=\left({135 + 273} \right)\;{\text{K}}\\&=408\;{\text{K}}\\\end{aligned}$

Also, the volume is to be converted into L. The conversion factor for this is,

${\text{1 mL}}={\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}$

So the volume of nitrogen is calculated as follows:

$\begin{aligned}{\text{Volume}}\left({\text{L}}\right)&=\left({{\text{300 mL}}} \right)\left({\frac{{{{10}^{ - 3}}{\text{L}}}}{{{\text{1 mL}}}}}\right)\\&=0.3\;{\text{L}}\\\end{aligned}$

The pressure of nitrogen is 780 mm Hg.

The volume of nitrogen is 0.3 L.

The temperature of nitrogen is 408 K.

The universal gas constant is 62.36367 L mmHg/mol K.

Substitute these values in equation (2).

$\begin{aligned}{\text{n}}&=\frac{{\left({{\text{780 mm Hg}}}\right)\left({{\text{0}}{\text{.3 L}}} \right)}}{{\left({{\text{62}}{\text{.3637 L mm Hg/K mol}}}\right)\left( {{\text{408 K}}}\right)}}\\&={\text{0}}{\text{.009196 mol}}\\&\approx {\text{0}}{\text{.009 mol}}\\\end{aligned}$

According to Avogadro law, one mole of any substance contains  ${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}$ molecules.

So the number of molecules of nitrogen is calculated as follows:

$\begin{aligned}\text{Number of molecules of nitrogen}&=\left(0.009\text{ mol}\right)\left(\dfrac{6.022\times 10^{23}\text{molecules}}{1\text{mol}}\right)\\&=5.4198\times 10^{21}\text{ molecules}\\&\bf \approx5.4\times 10^{21}\text{\bf molecules} \end{aligned}$

Learn more:

1. Which statement is true for Boyle’s law: https://brainly.com/question/1158880

2. Calculation of volume of gas: https://brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: ideal gas, pressure, volume, absolute temperature, equation of state, hypothetical, universal gas constant, moles of nitrogen, 780 mm Hg, pressure of nitrogen, 780 mm Hg, volume of nitrogen, 0.3 L, temperature of nitrogen, 408 K, universal gas constant, 62.36367 L mmHg/mol K.