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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 N\(_2\) gas) inside the ball is originally at 20.0\(^\circ\)C and 2.00 atm. The ball's inside diameter is 23.9 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
The final temperature is 371 K, and the change in internal energy is approximately 93.6 J.

Step by step solution

01

Find Initial Parameters and Understand Variables

Identify the key variables and initial conditions. We have an original diameter of 23.9 cm, which gives an original volume. The initial conditions are 20.0°C (or 293.15 K) and 2.00 atm. The ball is compressed to 80% of its original volume.
02

Calculate Initial and Final Volumes

Convert the diameter to meters and use the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). First, find the initial volume \( V_i \), and then multiply by 0.80 to find the final volume \( V_f \).
03

Convert Temperature to Kelvin

Convert the initial temperature from Celsius to Kelvin. Use the formula: \( T_{i} = T_{\text{Celsius}} + 273.15 \).

As given, \( T_{i} = 293.15 \text{ K} \).

04

Use the Adiabatic Process Equation

For an adiabatic process, we use the relationship \( TV^{\gamma-1} = \text{constant} \) and for nitrogen \( \gamma = \frac{7}{5} = 1.4 \). Using this formula, solve for the final temperature \( T_f \).
05

Calculate Final Temperature after Compression

We use the equation \( T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1} \) to find:\[ T_f = T_i \left( \frac{V_i}{V_f} \right)^{\gamma-1} \]Plug in the known values to find \( T_f \).
06

Calculate Change in Internal Energy

Change in internal energy \( \Delta U \) is given by \( \Delta U = n C_v \Delta T \) where \( C_v = \frac{5}{2} R \) for nitrogen. Calculate the number of moles \( n \) using \( PV = nRT \) with initial conditions. Then find \( \Delta T = T_f - T_i \) and calculate \( \Delta U \).
07

Solve for Change in Internal Energy

With the final temperature \( T_f \) and the initial temperature \( T_i \), calculate \( \Delta T \). Then use \( \Delta U = n \cdot C_v \cdot \Delta T \) to find the change in internal energy.

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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 an essential concept in physics and chemistry, describing the behavior of an ideal gas. The equation for the Ideal Gas Law is given by \( PV = nRT \), where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume of the gas.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the ideal gas constant, approximately 8.314 J/(mol·K).
  • \( T \) is the temperature in Kelvin.
This law helps us understand the relationship between pressure, volume, temperature, and the amount of gas. It assumes no interaction between gas molecules and that the gas occupies a large volume compared to the size of its molecules.
This makes it incredibly useful for calculations involving gaseous systems like the one in the basketball problem, where the gas behaves ideally.
Change in Internal Energy
Internal energy is the total energy contained within a system due to molecular motion and molecular position. When the conditions of the system change, such as during an adiabatic compression, there is a change in internal energy, \( \Delta U \). In the context of an ideal gas, the change in internal energy can be calculated using the formula:
  • \( \Delta U = nC_v\Delta T \)
Where:
  • \( n \) is the number of moles of the gas.
  • \( C_v \) is the molar heat capacity at constant volume.
  • \( \Delta T \) is the change in temperature.
In an adiabatic process, like the compression of the basketball, no heat is exchanged, so the change in internal energy is equal to the work done on the gas. Calculating \( \Delta U \) allows us to understand the energy dynamics during such processes.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships and conversions between heat and other forms of energy. There are four fundamental laws:
  • Zeroeth Law: If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
  • First Law (Law of Energy Conservation): Energy cannot be created or destroyed, only transformed from one form to another. For an ideal gas, this translates to \( \Delta U = Q - W \), where \( Q \) is heat added to the system, \( W \) is work done by the system.
  • Second Law: Energy has quality as well as quantity, and real processes occur in the direction of decreasing energy quality — typically, heat flows spontaneously from hot to cold.
  • Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum.
In our scenario, the adiabatic process is an application of the first and second laws, emphasizing that no heat is transferred to or from the gas.
Nitrogen Gas Properties
Nitrogen gas, denoted as \( N_2 \), is a diatomic molecule and makes up about 78% of Earth's atmosphere. It is odorless, colorless, and non-reactive under most conditions. In physics, when considering nitrogen as an ideal gas, several specific properties are considered:
  • It's diatomic, meaning it consists of two nitrogen atoms.
  • It's nonpolar, contributing to its stable nature at room temperature.
  • The molar heat capacity at constant volume, \( C_v \), for nitrogen is \( \frac{5}{2}R \) because it is a diatomic molecule, accommodating rotational energy at lower temperatures but not vibrational energy.
Understanding these properties helps in solving problems related to nitrogen gas, such as calculating its behavior under varying conditions in thermodynamic processes, like in the problem involving the basketball's adiabatic compression.

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Most popular questions from this chapter

The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine's cylinders to react chemically with gasoline. Many cars have a \(turbocharger\), which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an \(intercooler\) in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa), density \(\rho\) = 1.23 kg/m\(^3\), and temperature 15.0\(^\circ\)C. It is compressed adiabatically to 1.45 \(\times\) 10\(^5\) Pa. In the intercooler, the air is cooled to the original temperature of 15.0\(^\circ\)C at a constant pressure of 1.45 \(\times\) 10\(^5\) Pa. (a) Draw a \(pV\)-diagram for this sequence of processes. (b) If the volume of one of the engine's cylinders is 575 cm\(^3\), what mass of air exiting from the intercooler will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger alone?

Two moles of an ideal monatomic gas go through the cycle \(abc\). For the complete cycle, 800 J of heat flows out of the gas. Process \(ab\) is at constant pressure, and process \(bc\) is at constant volume. States \(a\) and \(b\) have temperatures \(T_a\) = 200 K and \(T_b\) = 300 K. (a) Sketch the \(pV\)-diagram for the cycle. (b) What is the work \(W\) for the process \(ca\)?

A gas in a cylinder expands from a volume of 0.110 m\(^3\) to 0.320 m\(^3\). Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 \(\times\) 10\(^5\) Pa during the expansion. The total heat added is 1.15 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

A gas in a cylinder is held at a constant pressure of 1.80 \(\times\) 10\(^5\) \(Pa\) and is cooled and compressed from 1.70 m\(^3\) to 1.20 m\(^3\). The internal energy of the gas decreases by 1.40 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the absolute value of the heat flow, [\(Q\)] , into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

A cylinder with a piston contains 0.250 mol of oxygen at 2.40 \(\times\) 10\(^5\) Pa and 355 K. The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a \(pV\)-diagram. Compute (b) the temperature during the isothermal compression; (c) the maximum pressure; (d) the total work done by the piston on the gas during the series of processes.

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