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\(\bullet$$\bullet\) 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 \(20.2^{\circ} \mathrm{C}\) and a pressure of 2.00 atm. The ball's diameter is 23.9 \(\mathrm{cm} .\) (a) What temperature does the air in the ball reach at its maximum compression? (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 temperature reaches 234.68 K. b) The internal energy decreases by 723.23 J.

Step by step solution

01

Convert Temperature to Kelvin

The initial temperature of the air is given as 20.2°C. Convert this temperature to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \] So, \[ T_1 = 20.2 + 273.15 = 293.35 \text{ K} \]
02

Calculate Initial Volume

The diameter of the basketball is given as 23.9 cm. The volume of a sphere is calculated using the formula:\[ V = \frac{4}{3} \pi r^3 \]The radius \( r \) is half of the diameter:\[ r = \frac{23.9}{2} = 11.95 \text{ cm} \]Convert to meters:\[ r = 0.1195 \text{ m} \]Thus, the initial volume \( V_1 \) is:\[ V_1 = \frac{4}{3} \pi (0.1195)^3 = 0.007158 \text{ m}^3 \]
03

Calculate Compressed Volume

When the ball is compressed, the volume reduces to 80% of the original. Thus, the compressed volume \( V_2 \) is:\[ V_2 = 0.8 \times V_1 = 0.8 \times 0.007158 = 0.005726 \text{ m}^3 \]
04

Apply Ideal Gas Law for Temperature Change

Assuming the gas is ideal, we can use the Ideal Gas Law in the form:\[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \]Given that pressure remains constant,\[ V_1 / T_1 = V_2 / T_2 \]Solving for \( T_2 \):\[ T_2 = T_1 \times \frac{V_2}{V_1} \]\[ T_2 = 293.35 \times \frac{0.005726}{0.007158} = 234.68 \text{ K} \]
05

Calculate Change in Internal Energy

The change in internal energy \( \Delta U \) for an ideal gas is given by:\[ \Delta U = n C_v \Delta T \]We need \( n \), the number of moles, which can be found using the ideal gas law initially:\[ PV = nRT \Rightarrow n = \frac{P_1 V_1}{RT_1} \]Where \( P = 2.00 \times 101325 \text{ Pa} \), and \( R = 8.314 \text{ J/mol K} \):\[ n \approx \frac{2.00 \times 101325 \times 0.007158}{8.314 \times 293.35} = 0.591 \text{ moles} \]The molar heat capacity for nitrogen \( C_v \approx 20.8 \text{ J/mol K} \).\[ \Delta T = T_2 - T_1 = 234.68 - 293.35 = -58.67 \text{ K} \]\[ \Delta U = 0.591 \times 20.8 \times (-58.67) \approx -723.23 \text{ J} \]
06

Interpret Results

The negative sign in \( \Delta U \) indicates that the internal energy decreases as the basketball compresses because energy is effectively being used to compress the gas.

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Key Concepts

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

Thermodynamics and the Ideal Gas Law
Thermodynamics is a fundamental branch of physics that deals with the relationships between heat, work, temperature, and energy. It provides a framework to understand how energy is transferred and transformed within a system. In this scenario, we tackle the thermodynamic concept using the ideal gas law, which is pivotal. The ideal gas law is expressed as \( PV = nRT \), linking pressure \( P \), volume \( V \), number of moles \( n \), and temperature \( T \) of a gas. The constant \( R \) is the universal gas constant.

This law assumes a hypothetical gas composed of randomly moving particles that do not interact, fitting well in many scenarios like our basketball exercise. During compression to 80% of its original volume while keeping pressure constant, this principle allows us to predict the temperature drop by expressing the relationship as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). This simple change reveals how gas cools as it is compressed in a closed system.
  • The gas law emphasizes how interconnected volume and temperature are.
  • Understanding it helps with predictions in engineering and physical sciences.
Molar Heat Capacity
Molar heat capacity is a measure of the amount of heat required to change the temperature of one mole of a substance by one degree Celsius (or one Kelvin). In our exercise, the molar heat capacity at constant volume \( C_v \) for nitrogen, which the basketball is filled with, is essential. Nitrogen behaves closely like an ideal gas; hence its \( C_v \) is straightforward in calculations.

For nitrogen, \( C_v \approx 20.8 \text{ J/mol K} \). It signifies how much heat the gas can absorb without increasing its internal temperature significantly. The concept of molar heat capacity explains the energy change in a system when not doing work (such as expanding or contracting without friction).
  • Thermodynamic properties like \( C_v \) help predict how a substance behaves under changes in energy.
  • This concept is crucial in designing systems involving temperature changes, like engines and refrigerators.
Internal Energy of a Gas
Internal energy refers to the total energy contained within a system, associated with the random motion of molecules. In a thermal process involving gases, internal energy change \( \Delta U \) is central. It is quantified as \( \Delta U = n C_v \Delta T \), linking the energy change with the amount of gas \( n \), its molar heat capacity \( C_v \), and changes in temperature \( \Delta T \).

When the basketball's air is compressed, calculating \( \Delta U \) indicates whether energy was absorbed or released. Notably, a negative \( \Delta U \) as seen here, implies a decrease in internal energy. It tells that the system lost energy, consistent with the cooling during volume reduction.
  • Understanding internal energy is vital for insights into energy conservation in physical processes.
  • It aids in determining overall system efficiency in heat engines and similar devices.
Temperature Conversion Techniques
Temperature conversion is a basic but crucial step in thermodynamics that ensures consistency and accuracy in calculations. We often need temperatures in Kelvin for thermodynamic equations, given its absolute nature (no negatives) aligning with energy processes best.

For instance, converting degrees Celsius to Kelvin involves a simple addition: \[ T(K) = T(°C) + 273.15 \]. This step was the initial move in our problem, converting 20.2°C into 293.35 K. This uniformity is key, as Kelvin is the standard unit in thermodynamics.
  • Always convert to Kelvin when working with thermodynamic equations.
  • This conversion prevents errors in calculating energy changes and other thermal properties.

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Most popular questions from this chapter

\(\bullet$$\bullet\) STP. The conditions of standard temperature and pressure (STP) are a temperature of \(0.00^{\circ} \mathrm{C},\) and a pressure of 1.00 atm. (a) How many liters does 1.00 mol of any ideal gas occupy at STP? (b) For a scientist on Venus, an absolute pressure of 1 Venusian-atmosphere is 92 Earth-atmospheres. Of course she would use the Venusian atmosphere to define STP. Assuming she kept the same temperature, how many liters would 1 mole of ideal gas occupy on Venus?

As the temperature of a monatomic gas increases, we expect its specific heat to A. remain the same B. increase C. decrease

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