Question

Modern vacuum pumps make it easy to attain pressures of the order of 10−13 atm in the laboratory. consider a 1.0 cm3 volume of air and treat the air as an ideal gas. (a) at a pressure of 9.00 × 10−14 atm and an ordinary temperature of 273.15 k (0.0 ◦c), how many molecules are present in a volume of 1.00 cm3 ? (b) how many molecules would be present at the same temperature but at 1.00 atm instead? this is known as loschmidt’s number.

Answer 1

Since the gases behave ideally, we use the ideal gas equation:

PV = nRT
where
P is pressure
V is volume
n is the number of moles
R is the gas constant equal to 0.0821 L·atm/mol·K
T is the absolute temperature

a.) P = 9.00×10⁻¹⁴ atm; T = 273.15 K; V = 1 cm³ = 0.001 L

(9.00×10⁻¹⁴ atm)(0.001 L) = n(0.0821 L·atm/mol·K)(273.15 K)
n = 4.013×10⁻¹⁸ moles
Since there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 2,416,847

b.) P = 1 atm; T = 273.15 K; V = 1 cm³ = 0.001 L

(1 atm)(0.001 L) = n(0.0821 L·atm/mol·K)(273.15 K)
n = 4.459×10⁻⁶ moles
Since there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 2.685×10¹⁹

Answer 2

(a) $\boxed{{\text{2416790 molecules}}}$ are present in ${\text{1 c}}{{\text{m}}^3}$ of air.

(b) $\boxed{{\text{2}}{\text{.68}} \times {\text{1}}{{\text{0}}^{{\text{19}}}}{\text{ molecules}}}$ are present at 273.15 K and 1 atm.

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 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 the gas mixture.

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

(a)

The volume of air is to be converted into L. The conversion factor for this is,

${\text{1 c}}{{\text{m}}^3}={10^{-3}}{\text{ L}}$

So the volume of air can be calculated as follows:

$\begin{aligned}{\text{Volume of air}}&=\left({{\text{1 c}}{{\text{m}}^3}} \right)\left( {\frac{{{{10}^{-3}}{\text{ L}}}}{{{\text{1 c}}{{\text{m}}^3}}}}\right)\\&=0.001{\text{ L}}\\\end{aligned}$

The pressure of air is $9 \times {10^{ - 14}}\;{\text{atm}}$.

The volume of air is 0.001 L.

The temperature of air is 273.15 K.

The universal gas constant is 0.0821 L atm/K mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{n}}&=\frac{{\left({{\text{9}}\times{\text{1}}{{\text{0}}^{-14}}{\text{ atm}}}\right)\left({0.001{\text{ L}}} \right)}}{{\left({0.0821{\text{ L atm/K mol}}} \right)\left( {27{\text{3}}{\text{.15 K}}} \right)}}\\&=4.01326 \times {10^{ - 18}}{\text{ mol}}\\\end{aligned}$

Avogadro’s law states that one mole of the substance contains ${\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}$ molecules.

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

$\begin{aligned}{\text{Number of molecules of air}}{\mathbf{=}}\left( {4.01326 \times {{10}^{ - 18}}{\text{ mol}}}\right)\left( {\frac{{{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}}\right)\\={\text{2416789}}{\text{.905}}\;{\text{molecules}}\\\approx{\text{2416790 molecules}}\\ \end{aligned}$

(b)

The pressure of air is 1 atm.

The volume of air is 0.001 L.

The temperature of air is 273.15 K.

The universal gas constant is 0.0821 L atm/K mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{n}}&=\frac{{\left( {1{\text{ atm}}} \right)\left( {0.001{\text{ L}}} \right)}}{{\left( {0.0821{\text{ L atm/K mol}}} \right)\left( {27{\text{3}}{\text{.15 K}}} \right)}}\\&=0.0000445918{\text{ mol}} \\ \end{aligned}$

The number of molecules of air is calculated as follows:

$\begin{aligned}{\text{Number of molecules of air}}{&\mathbf{ = }}\left( {0.0000445918{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}}\right)\\ &={\text{2}}{\text{.68322}} \times {\text{1}}{{\text{0}}^{{\text{19}}}}\;{\text{molecules}}\\&\approx {\text{2}}{\text{.68}} \times {\text{1}}{{\text{0}}^{{\text{19}}}}{\text{ molecules}} \\ \end{aligned}$

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: P, V, n, R, T, air, 273.15 K, 1 atm, ideal gas, ideal gas equation, 0.0000445918 mol, 0.001 L, 2.68*10^19 molecules, 2416790 molecules, pressure, volume, temperature, Avogadro’s law.