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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

A large research balloon containing \(2.00 \times 10^{3} \mathrm{~m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. \(\mathbf{P} 19.50\) ). Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

Nitrogen gas in an expandable container is cooled from 50.0\(^\circ\)C to 10.0\(^\circ\)C with the pressure held constant at 3.00 \(\times\) 10\(^5\) Pa. The total heat liberated by the gas is 2.50 \(\times\) 10\(^4\) J. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

On a warm summer day, a large mass of air (atmospheric pressure 1.01 \(\times\) 10\(^5\) Pa) is heated by the ground to 26.0\(^\circ\)C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 \(\times\) 10\(^5\) Pa. Assume that air is an ideal gas, with \(\Upsilon\) = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C\(^\circ\) per 100 m of altitude, is called the dry \(adiabatic\) \(lapse\) \(rate\).)

In another test, the valve of a 500-L cylinder full of the gas mixture at 2000 psi (gauge pressure) is opened wide so that the gas rushes out of the cylinder very rapidly. Why might some \(N_2O\) condense during this process? (a) This is an isochoric process in which the pressure decreases, so the temperature also decreases. (b) Because of the rapid expansion, heat is removed from the system, so the internal energy and temperature of the gas decrease. (c) This is an isobaric process, so as the volume increases, the temperature decreases proportionally. (d) With the rapid expansion, the expanding gas does work with no heat input, so the internal energy and temperature of the gas decrease.

Five moles of an ideal monatomic gas with an initial temperature of 127\(^\circ\)C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?

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