Question

a. How many molecules strike a \(1.00 \mathrm{cm}_{2}\) surface during 1 minute if the surface is exposed to \(\mathrm{O}_{2}\) at 1 atm and \(298 \mathrm{K} ?\) b. Ultrahigh vacuum studies typically employ pressures on the order of \(10^{-10}\) Torr. How many collisions will occur at this pressure at \(298 \mathrm{K} ?\)

Short answer

a. Under the given conditions, the number of \(\mathrm{O}_2\) molecules that strike a \(1.00 \mathrm{cm}^2\) surface per minute is approximately \(7.36 \times 10^{21}\) molecules. b. The number of collisions per minute when exposed to a pressure of \(10^{-10} \mathrm{Torr}\) and at \(298 \mathrm{K}\) is approximately \(5.67 \times 10^7\) collisions.

Step-by-step solution

Convert the given data into the SI Units

First, let's convert the given surface area and time into the SI units. Surface area = \(1.00 \mathrm{cm}^2 = 1.00 \times 10^{-4} \mathrm{m}^2\) Time = 1 minute = 60 seconds.

Use ideal gas law equation to calculate the number of molecules

Using the ideal gas law equation, we can calculate the number of \(\mathrm{O}_2\) molecules: \(PV = nRT\) Where: P = pressure = 1 atm = \(1.013 \times 10^5 \mathrm{Pa}\) (using the conversion 1 atm = 101.3 kPa), V = volume, n = number of moles, R = universal gas constant = 8.314 J/(mol K), T = temperature = 298 K. First, let's write the number of moles (n) in terms of total mass (m) and molar mass (M). This is important since we have to find the number of molecules striking the surface. \(n = \frac{m}{M}\) Now, we substitute n in the ideal gas law equation: \(PV = \frac{m}{M}RT\)

Calculate the total mass of O2

Rearrange the equation to find the mass (m) of \(\mathrm{O}_2\): \(m = \frac{PMV}{RT}\) Now, we need to find the volume (V). To do that, we have to use the given surface area (1.00 cm^2). Let's assume the volume of the space is a cylinder with a height of 1 m (it's a rough assumption): \(V = A \cdot h = 1.00 \times 10^{-4} \times 1.00 = 1.00 \times 10^{-4} \mathrm{m}^3\) Now we can calculate the mass (m): \(m = \frac{(1.013 \times 10^5)(1.00 \times 10^{-4})(298)}{8.314(298)}\) \(m = 0.039 \mathrm{g}\)

Calculate the number of O2 molecules using Avogadro's number

Now we have the total mass of \(\mathrm{O}_2\), and we can calculate the total number of molecules: \(N = \frac{m}{M} \times N_A\) Where: N = total number of \(\mathrm{O}_2\) molecules, M = molar mass of \(\mathrm{O}_2\) = 32 g/mol (approximately), \(N_A\) = Avogadro's number = \(6.022 \times 10^{23}\) molecules/mol. So, the number of \(\mathrm{O}_2\) molecules in the volume is: \(N = \frac{0.039}{32} \times 6.022 \times 10^{23} = 7.36 \times 10^{21} \mathrm{molecules}\) **Part B**

Convert pressure from Torr to atm

First, let's convert the given pressure to the SI unit (Pa): \(P = 10^{-10} \mathrm{Torr} \times \frac{1 \mathrm{atm}}{760 \mathrm{Torr}} \times \frac{1.013 \times 10^5 \mathrm{Pa}}{1 \mathrm{atm}} = 1.333 \times 10^{-13} \mathrm{Pa}\)

Derive the number of collisions from the ideal gas law equation

Now, let's use the ideal gas law equation with the new pressure to derive the total number of molecules and collisions for this pressure and temperature: \(PV = nRT\) \(PV = (\frac{m}{M}) RT\) Where n, R, and T remain the same as before.

Calculate the mass using the new pressure

Rearranging the equation to find the mass m: \(m = \frac{PMV}{RT}\) And using the same volume (as it was assumed in Part A) to find the mass m for the new pressure: \(m = \frac{(1.333 \times 10^{-13})(1.00 \times 10^{-4})(298)}{8.314(298)}\) \(m = 5.03 \times 10^{-18} \mathrm{g}\)

Calculate the number of collisions at the new pressure

Using the same method as before to find the number of \(\mathrm{O}_2\) molecules: \(N = \frac{m}{M} \times N_A\) \(N = \frac{5.03 \times 10^{-18}}{32} \times 6.022 \times 10^{23} = 9.45 \times 10^{5} \mathrm{molecules}\) Since we are considering one minute, or 60 seconds, we just need to multiply the number of molecules by 60 to get the number of collisions. Number of collisions = \(9.45 \times 10^5 \times 60 = 5.67 \times 10^7\) So, the number of collisions per minute at the pressure of \(10^{-10} \mathrm{Torr}\) and at 298 K is approximately \(5.67 \times 10^7\) collisions.

Key concepts

Understanding Molecular Collisions

Molecular collisions refer to the interactions between molecules in a gas, typically occurring when they bump into one another due to their random motion. In gases, these collisions are frequent, as molecules are constantly moving in all directions. The behavior of molecules in a gas is largely defined by what is known as the kinetic molecular theory. This theory suggests that the gas is made up of many small particles that are in constant, random motion. As they collide, energy can be transferred between molecules, affecting their speed and direction.

In the context of the ideal gas law, these collisions play a critical role. At a molecular level, when a gas is at a higher temperature, the molecules move more vigorously, leading to more frequent and more intense collisions with surfaces. On the other hand, lower temperatures result in slower molecular movement and fewer collisions.

Understanding molecular collisions is essential when calculating how many molecules strike a given surface area within a specific time period. The frequency and impact of collisions are influenced by factors such as pressure, temperature, and volume, often needing complex equations like the ideal gas law to describe their behavior accurately.

Exploring Avogadro's Number

Avogadro's number is a fundamental constant in chemistry that defines the number of atoms or molecules in one mole of a substance. This number is typically expressed as \(6.022 \times 10^{23}\), meaning for every mole of a substance, there are that many individual atoms or molecules. This concept is pivotal in molecular chemistry, as it allows chemists to scale atoms and molecules from a microscopic level to macroscopic quantities that can be observed and measured.

In practical terms, Avogadro's number helps us convert the mass of a substance to the number of particles it contains, which is particularly useful when analyzing chemical reactions and the behavior of gases. In exercises dealing with gases, such as the ones governed by the ideal gas law, Avogadro's number allows us to determine the number of molecules present in a gas sample by connecting it with the number of moles. For instance, knowing the total mass of gas and its molar mass, we can use Avogadro's number to calculate the total number of molecules and hence the number of molecular collisions as mentioned in the exercise.

Overall, Avogadro's number serves as a vital link between the macroscopic measurements we can make and the microscopic scale at which atoms and molecules operate.

The Importance of Pressure Conversion

Pressure conversion is an important skill when working with gases, as pressure is often given in various units. Common units include atmospheres (atm), pascals (Pa), and torr. Being able to convert between these units is essential for applying the ideal gas law correctly. The ideal gas law, expressed as \(PV = nRT\), requires pressure in one consistent unit to perform calculations accurately.

For example, 1 atm is equivalent to \(1.013 \times 10^5\) pascals (Pa) and approximately 760 torr. In everyday gas law problems, you may need to convert pressure from these units to ensure consistency with other quantities. This is crucial for calculating the number of molecules or predicting the behavior of gases under different conditions.

Converting pressure correctly becomes particularly critical in applications like ultrahigh vacuum studies, where pressures significantly lower than atmospheric pressure (like \(10^{-10}\) torr) are used. Here, getting the conversion to the SI unit (Pa) right is necessary to determine the number of molecular collisions or the behavior of gases in such low-pressure environments. Mastering pressure conversion thus allows students and professionals to better interpret and predict gas behavior in various scientific and real-world scenarios.