/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A player bounces a basketball on... [FREE SOLUTION] | 91Ó°ÊÓ

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A player bounces a basketball on the floor, compressing it to 80.0\(\%\) of its original volume. The air (assume it is essentially \(\mathrm{N}_{2}\) gas) inside the ball is originally at a temperature of \(20.0^{\circ} \mathrm{C}\) and a pressure of 2.00 atm. The ball's inside diameter is 23.9 \(\mathrm{cm}\) . (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal. (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

Short Answer

Expert verified
(a) The air temperature reaches approximately 312.47 K. (b) The change in internal energy is about 29.59 J.

Step by step solution

01

Convert Initial Information into Standard Units

First, we need to convert all given data into standard units. The initial temperature of the gas is given as \(20.0^{\circ}\mathrm{C}\), which converts to Kelvin using the formula \(T(K) = T(^{\circ}C) + 273.15\). So, the initial temperature is \(293.15\, K\). The original pressure of the gas is \(2.00\, atm\), which is approximately \(2.00 \times 101325\, Pa = 202650\, Pa\). The diameter of the basketball is \(23.9\, cm\), which is equivalent to \(0.239\, m\).
02

Calculate Original Volume of the Gas

Using the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius of the ball. The radius in meters is half the diameter, so \(r = 0.239/2\, m = 0.1195\, m\).Thus, the original volume \(V_{1}\) is \(\frac{4}{3} \pi (0.1195)^{3} \approx 0.00715\, m^3\).
03

Determine Compressed Volume of the Gas

When compressed to 80\% of its original volume, the new volume \(V_2\) will be \(80\%\) of \(0.00715\, m^3\).Calculate \(V_2 = 0.80 \times 0.00715\, m^3 \approx 0.00572\, m^3\).
04

Apply Adiabatic Process to Find New Temperature

For an adiabatic process of an ideal gas, \(PV^\gamma = \text{constant}\) and \(T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}\). For nitrogen (\(N_2\)), \(\gamma \approx 1.4\).Calculate the new temperature \(T_2 = 293.15 \left( \frac{0.00715}{0.00572} \right)^{0.4} \approx 312.47\, K\).
05

Calculate Change in Internal Energy

The change in internal energy for a diatomic ideal gas like \(N_2\) can be found using \(\Delta U = nC_v(T_2 - T_1)\), where \(C_v = \frac{5}{2}R\). First, find \(n\) using \(PV = nRT\), with the initial conditions: \(n = \frac{202650 \times 0.00715}{8.314 \times 293.15} \approx 0.0595\, mol\).Then, \(\Delta U = 0.0595 \times \frac{5}{2} \times 8.314 \times (312.47 - 293.15) \approx 29.59\, J\).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas. It's given by the equation \( PV = nRT \), where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume it occupies.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the ideal gas constant (8.314 J/(mol·K)).
  • \( T \) is the temperature in Kelvin.
This equation shows how pressure, volume, and temperature are interrelated for a given amount of gas. It's useful for solving various problems, such as calculating the number of moles in a gas at a particular temperature and pressure or finding the volume occupied by a gas when conditions change. In our basketball example, the Ideal Gas Law helped determine the number of moles of air in the ball, critical for later calculating the change in internal energy during compression.
Thermodynamics
Thermodynamics is the branch of physics concerned with heat, temperature, and their relation to energy and work. It encompasses principles that describe how energy is transferred and transformed. A key concept in thermodynamics is the adiabatic process, where no heat is transferred into or out of the system.
In the context of a bouncing basketball, when you compress it quickly, the process is considered adiabatic. During such a process:
  • The internal energy of the system changes, but it does so without heat exchange with the environment.
  • For an ideal gas, this results in a change in temperature and pressure directly linked to changes in volume.
Overall, understanding thermodynamics allows us to predict how temperature will increase when a gas, like the air in the basketball, is compressed quickly without allowing it to exchange heat with its surroundings.
Change in Internal Energy
The change in internal energy \( \Delta U \) of a system is a core concept in thermodynamics, especially involving ideal gases. It's calculated using the equation \( \Delta U = nC_v(T_2 - T_1) \), where:
  • \( n \) is the number of moles of gas.
  • \( C_v \) is the molar heat capacity at constant volume.
  • \( T_2 \) and \( T_1 \) are the final and initial temperatures, respectively.
In this basketball scenario, the internal energy change is calculated based on the difference in temperature due to compression. Since the compression is adiabatic, the increase in pressure leads to a rise in temperature. By using the earlier derived number of moles and knowing \( C_v \) for nitrogen, which is typically \( \frac{5}{2}R \), the change in internal energy was determined. In our problem, this resulted in an increase in internal energy, indicating the system did work while energy was transformed within.

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Most popular questions from this chapter

A quantity of air is taken from state \(a\) to state \(b\) along a path that is a straight line in the \(p V\) -diagram (Fig. Pl9.39). (a) In this process, does the temperature of the gas increase, decrease, or stay the same? Explain. (b) If \(V_{a}=0.0700 \mathrm{m}^{3}\) , \(V_{b}=0.1100 \mathrm{m}^{3}, \quad p_{a}=1.00 \mathrm{x}\) \(10^{5} \mathrm{Pa},\) and \(p_{b}=1.40 \times 10^{5}\) \(\mathrm{Pa},\) what is the work \(W\) done by that gas in this process? Assume that the gas may be treated as ideal.

During an isothermal compression of an ideal gas, 335 \(\mathrm{J}\) of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

Two moles of an ideal gas are heated at constant pressure from \(T=27^{\circ} \mathrm{C}\) to \(T=107^{\circ} \mathrm{C}\) (a) Draw a p \(V\) -diagram for this process. (b) Calculate the work done by the gas.

CP Oscillations of a Piston. A vertical cylinder of radius \(r\) contains a quantity of ideal gas and is fitted with a piston with mass \(m\) that is free to move (Fig. P19.69). The piston and the walls of the cylinder are frictionless, and the entire cylinder is placed in a constant-temperature bath. The outside air pressure is \(p_{0} .\) In equilibrium, the piston sits at a height \(h\) above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance \(h+y\) above the bottom of the cylinder, where \(y\) is much less than \(h\) . (c) After the piston is displaced from equilibrium and released, it oscillates up and down. Find the frequency of these small oscillations. If the displacement is not small, are the oscillations simple harmonic? How can you tell?

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