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In a period of \(1.0 \mathrm{~s}, 5.0 \times 10^{23}\) nitrogen molecules strike a wall of area \(8.0 \mathrm{~cm}^{2}\). If the molecules move at \(300 \mathrm{~m} / \mathrm{s}\) and suike the wall head on in a perfectly elastic collision, find the pressure exerted on the wall. (The mass of one \(\mathrm{N}_{2}\) molecule is \(\left.4.68 \times 10^{-26} \mathrm{~kg} .\right)\)

Short Answer

Expert verified
The pressure exerted on the wall by the nitrogen molecules is \(1.755 \times 10^{6}\) Pa.

Step by step solution

01

Calculate the Change in Momentum of a Single Nitrogen Molecule

As this is a perfectly elastic collision, the molecules will bounce back with the same speed they came in. Therefore, the change in momentum \(\Delta p\) per molecule will be twice the initial momentum. Each molecule's initial momentum \(p_{initial}\) can be calculated from its mass \(m\) and velocity \(v\), using the formula \(p_{initial} = m*v\). Substituting \(m = 4.68 \times 10^{-26} \mathrm{~kg}\) and \(v = 300 \mathrm{~m/s}\), we get \(p_{initial} = 4.68 \times 10^{-26} \mathrm{~kg} * 300 \mathrm{~m/s} = 1.404 \times 10^{-23} \mathrm{~kg m/s}\). Thus, \(\Delta p = 2p_{initial} = 2.808 \times 10^{-23} \mathrm{~kg m/s}\).
02

Determine the Total Impulse Delivered to the Wall

Next, we need to multiply the change in momentum per molecule by the total number of molecules striking the wall per second to get the total impulse delivered to the wall per second. Given \(5.0 \times 10^{23}\) molecules striking the wall per second, the total impulse is \(I_{total} = 5.0 \times 10^{23} * \Delta p = 5.0 \times 10^{23} * 2.808 \times 10^{-23} \mathrm{~kg m/s} = 1404 \mathrm{~kg m/s}\).
03

Compute the Force Exerted on the Wall

Force is then calculated from the total impulse delivered to the wall per second using the formula \(F = I_{total}/t\). With \(I_{total} = 1404 \mathrm{~kg m/s}\) and \(t = 1.0s\), we get \(F = 1404 \mathrm{~kg m/s}/1.0s = 1404 \mathrm{~N}\).
04

Calculate the Pressure

Finally, pressure is calculated from the force and the area of the wall. The area given in cm\u00b2 needs to be converted to m\u00b2, yielding \(A = 8.0 \mathrm{~cm}^{2} * (1 \mathrm{~m}/100 \mathrm{~cm})^{2} = 0.0008 \mathrm{~m}^{2}\). Using the formula \(P = F/A\), pressure is then \(P = 1404 \mathrm{~N}/0.0008 \mathrm{~m}^{2} = 1.755 \times 10^{6} \mathrm{~Pa}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Collision
An elastic collision is a type of collision where the total kinetic energy is conserved. In fluid dynamics and statistical mechanics, we're often interested in how gas molecules collide and interact. During an elastic collision, each molecule retains its energy and simply changes direction.

In the context of our problem, we considered nitrogen molecules colliding with a wall head-on. These molecules are said to undergo perfectly elastic collisions, meaning they bounce back with the same velocity upon hitting the wall. This maintains the system's energy level without any energy loss. In practical terms, for a perfectly elastic collision, the velocity before and after the impact is the same, which directly affects the momentum calculations involved.
Momentum Change
Momentum is a vector quantity defined as the product of an object's mass and its velocity. In our exercise, the concept of momentum change is central when analyzing collisions. When a nitrogen molecule collides with the wall, its momentum changes because its direction changes. However, the magnitude of the velocity remains the same due to the elastic nature of the collision.

The change in momentum, \(\Delta p\), for a molecule is twice the initial momentum because in a perfectly elastic collision, a molecule like nitrogen that hits a wall and bounces back reverses its velocity, and thus its momentum doubles. Mathematically, if \( p_{initial} \) is the initial momentum calculated as \(m \times v\), the momentum change would be \(\Delta p = 2 \times p_{initial}\). This change in momentum is crucial for calculating the total impulse delivered to the wall.
Impulse Delivery
Impulse is the product of force and the time period it acts upon an object, or equivalently, the change in momentum of the object. In our scenario, impulse is used to determine how the collision dynamics translate into pressure on the wall.

Given that \(5.0 \times 10^{23}\) nitrogen molecules strike the wall every second, the total impulse is calculated by multiplying the change in momentum of one molecule by the number of molecules. This calculation gives us the total impulse per second against the wall. Impulse is denoted in units of \(\text{kg m/s}\), and it's a crucial stepping stone in calculating the force exerted by the molecules, leading us to further derive the pressure.
Kinetic Theory of Gases
The kinetic theory of gases is a scientific theory that explains gas properties in terms of molecular motion. This theory helps us understand the behavior of gas particles through various principles, like how they move, collide, and exert pressure.

In our exercise, this theory is pivotal because it provides the fundamental explanation for why and how the molecules exert pressure on the walls. According to the kinetic theory, gas molecules are in constant random motion, and when they collide with a surface like the wall in our problem, they exert force. Collectively, these countless collisions result in measurable pressure. The theory assumes ideal conditions, where molecules do not exert forces on each other except during elastic collisions, enhancing our understanding of pressure and kinetic energy considerations in gas dynamics.

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Most popular questions from this chapter

Two small containers, each with a volume of \(100 \mathrm{~cm}^{3}\), contain helium gas at \(0^{-} \mathrm{C}\) and \(1.00\) atm pressure. The two containers are joined by a small open tube of negligible volume, allowing gas to flow from one container to the other. What common pressure will exist in the two containers if the temperature of one container is raised to \(100^{\circ} \mathrm{C}\) while the other container is kept at \(0^{\circ} \mathrm{C}\) ?

A solid substance has a density \(\rho_{0}\) at a temperature \(T_{0}\). If its temperature is increased by an amount \(\Delta T\), show that its density at the higher temperature is given by $$ \rho=\frac{\rho_{0}}{1+\beta \Delta T} $$

What is the average kinetic energy of a molecule of oxygen at a temperature of \(300 \mathrm{~K}\) ?

The average coefficient of volume expansion for carbon tetrachloride is \(5.81 \times 10^{-4}\left({ }^{\circ} \mathrm{C}\right)^{-1}\). If a \(50.0\) -gal steel container is filled completely with carbon tetrachloride when the temperature is \(10.0^{\circ} \mathrm{C}\), how much will spill over when the temperature rises to \(30.0^{\circ} \mathrm{C}\) ?

A vertical cylinder of crosssectional area \(A\) is fitted with \(\underline{a}\) tight-fitting, frictionless piston of mass \(m\) (Fig. P10.54). (a) If \(n\) moles of an ideal gas are in the cylinder at a temperature of \(T\), use Newton's second law for equilibrium to show that the height \(h\) at which the piston is in equilibrium under its own weight is given by $$ h=\frac{n R T}{m g+P_{n} A} $$ where \(P_{0}\) is atmospheric pressure. (b) Is the pressure inside the cylinder less than, equal to, or greater than atmospheric pressure? (c) If the gas in the cylinder is warmed, how would the answer for \(h\) be affected?

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