Question

Part b an "empty" container is not really empty if it contains air. how may moles of nitrogen are in an "empty" two-liter cola bottle at atmospheric pressure and room temperature (25∘c)? assume ideal behavior. express your answer to three significant figures and include the appropriate units.

Answer 1

6.38x10^-2 moles
       First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
   2 L / 24.465 L/mol = 0.081749438 mol
       Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
    0.081749438 mol * 0.78 = 0.063764562 mol.
        Rounding to 3 significant figures gives 6.38x10^-2 moles

Answer 2

$\boxed{{\text{0}}{\text{.0637 mol}}}$  of nitrogen are present in a two-litre cola bottle at atmospheric pressure and room temperature.

Further Explanation:

An ideal gas is a hypothetical gas that is composed of 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 the equation of state for any hypothetical gas. The expression for the ideal gas equation is as follows:

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

Here,

P is the pressure of air.

V is the volume of air.

T is the absolute temperature of the air.

n is the number of moles of air.

R is the universal gas constant.

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

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

The value of P is 1 atm.

The value of V is 2 L.

The value of R is 0.0821 L atm/K mol.

The value of T is 298.15 K.

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

${\text{1 }}^\circ {\text{C}} =273.1{\text{5 K}}$

So the temperature of the air can be calculated as follows:

$\begin{aligned}{\text{Temperature of gas}} &= \left( {25 + 273.15} \right)\;{\text{K}} \\&= {\text{298}}{\text{.15 K}}\\\end{aligned}$

Substitute these values in equation (2).

$\begin{aligned}{\text{n}}&=\frac{{\left( {1{\text{ atm}}}\right)\left( {2{\text{ L}}} \right)}}{{\left( {0.0821{\text{ L atm/K mol}}} \right)\left( {298.{\text{15 K}}} \right)}}\\&= 0.0817{\text{ mol}}\\\end{aligned}$

The number of moles of air comes out to be 0.0817 mol. But it does not contain nitrogen only. Nitrogen is 78 % of air.

The amount of nitrogen in the air can be calculated as follows:

$\begin{aligned}{\text{Amount of }}{{\text{N}}_{\text{2}}} &= \left( {{\text{0}}{\text{.0817 mol}}}\right)\left({\frac{{78\;\% }}{{100\;\% }}}\right)\\&= 0.0{\text{637 mol}}\\\end{aligned}$

Therefore 0.0637 moles of nitrogen are present in the container.

Learn more:

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

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

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: P, V, n, R, T, air, 273.15 K, 1 atm, ideal gas, ideal gas equation, N2, air, 0.0817 mol, 78 %, 0.0637 mol, 298.15 K.