Problem 78 - University Physics with Modern Physics

Question

A closed auditorium of volume \(2.50 \cdot 10^{4} \mathrm{~m}^{3}\) is filled with 2000 people at the beginning of a show, and the air in the space is at a temperature of \(293 \mathrm{~K}\) and a pressure of \(1.013 \cdot 10^{5} \mathrm{~Pa}\). If there were no ventilation, by how much would the temperature of the air rise during the \(2.00-\mathrm{h}\) show if each person metabolizes at a rate of \(70.0 \mathrm{~W} ?\)

Step-by-Step Solution

Answer

Answer: The temperature of the air inside the auditorium would increase by approximately 4.7 K during the show without ventilation.

Step 1: Calculate the total heat produced during the show.

To determine the total heat produced during the show, we need to multiply the metabolic rate of each person by the number of people and the duration of the show in seconds. \(Q = \text{Number of people} \times \text{Metabolic rate} \times \text{Duration of the show in seconds}\) \(Q = 2000 \times 70.0 \, \text{W} \times 2 \, \text{h} \times 3600 \, \frac{\text{s}}{\text{h}}\) \(Q = 1004 \times 10^6 \, \text{J}\)

Step 2: Calculate the number of moles of air in the auditorium.

Using the ideal gas law, we can determine the number of moles of air inside the auditorium. \(PV = nRT\) We will rearrange the formula to find n: \(n = \frac{PV}{RT}\) We are given the initial pressure (\(P = 1.013 \times 10^5 \, \text{Pa}\)), volume (\(V = 2.50 \times 10^4 \, \text{m}^3\)), and temperature (\(T = 293 \, \text{K}\)) of the air, and the gas constant (\(R = 8.314 \, \frac{\text{J}}{\text{mol}\cdot\text{K}}\)). \(n = \frac{(1.013 \times 10^5 \, \text{Pa})(2.50 \times 10^4 \, \text{m}^3)}{(8.314 \, \frac{\text{J}}{\text{mol} \cdot \text{K}})(293 \, \text{K})}\) \(n \approx 1.03 \times 10^6 \, \text{mol}\)

Step 3: Apply the first law of thermodynamics.

We will now apply the first law of thermodynamics, which states: \(\Delta U = Q - W\) In this case, there is no work done on or by the system, so \(W = 0\). The change in internal energy is given by: \(\Delta U = n \, C_v \, \Delta T\) Where \(C_v\) is the molar heat capacity at constant volume for air, which is approximately \(20.8 \, \frac{\text{J}}{\text{mol} \cdot \text{K}}\). We can solve for the change in temperature: \(Q = n \, C_v \, \Delta T\) \(\Delta T = \frac{Q}{n \, C_v}\) \(\Delta T = \frac{1004 \times 10^6 \, \text{J}}{(1.03 \times 10^6 \, \text{mol})(20.8 \, \frac{\text{J}}{\text{mol} \cdot \text{K}})}\) \(\Delta T \approx 4.7 \, \text{K}\) The temperature of the air would rise by approximately \(4.7 \, \text{K}\) during the show without ventilation.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that helps to describe the behavior of gases. It combines several properties of gases into a simple mathematical relationship. The law is represented by the equation:

• \( PV = nRT \)

Where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (8.314 \( \, \text{J/mol} \, \cdot \, \text{K} \)),
• \( T \) is the temperature in Kelvin.

This law assumes that gases are made up of a large number of very small particles moving in random directions, and it applies ideally when these particles do not interact with each other except through elastic collisions. In the given problem, we use the Ideal Gas Law to compute the number of moles of air in the auditorium. By knowing the pressure, volume, and temperature, the number of moles \( n \) can be calculated. This forms the basis to further calculate changes in the thermodynamic properties of the gas.

First Law of Thermodynamics

The First Law of Thermodynamics is one of the essential principles that governs energy conservation. It is expressed mathematically as:

• \( \Delta U = Q - W \)

Where:

• \( \Delta U \) is the change in internal energy of the system,
• \( Q \) is the heat added to the system,
• \( W \) is the work done by the system.

For the auditorium exercise, we assume that no work is done (i.e., \( W = 0 \)), simplifying the equation to \( \Delta U = Q \). The heat produced by the occupants increases the internal energy, leading to a rise in temperature. This principle helps us understand how energy is transformed and how it affects the system's temperature.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another. It occurs through three main mechanisms: conduction, convection, and radiation. In the auditorium problem, heat is generated internally by the body's metabolism of people. This heat raises the air temperature inside the closed space:

• **Conduction** is the transfer of heat through a material. In this case, it is minimal without direct contact.
• **Convection** involves the movement of air itself distributing the heat internally, which is likely in a crowded room.
• **Radiation** can be present as the heat radiates from bodies to surrounding air.

Due to the closed environment, much of the generated heat stays within the space, leading to an incremental rise in temperature, calculated using thermodynamic principles.

Molar Heat Capacity

Molar Heat Capacity is an important concept that tells us how much heat is needed to increase the temperature of a mole of a substance by one degree Kelvin. For a gas at constant volume, this is denoted as \( C_v \). In our problem, the value of \( C_v \) for air is approximately \( 20.8 \, \text{J/mol} \, \cdot \, \text{K} \).

• This value is used to determine how much the temperature of the air inside the auditorium would rise when a certain amount of heat is added.
• The formula used is: \( \Delta T = \frac{Q}{n \, C_v} \), where \( \Delta T \) is the change in temperature, \( Q \) is the heat added, \( n \) is the number of moles, and \( C_v \) is the molar heat capacity.

This understanding allows us in practical scenarios to estimate how energy affects temperature change, crucial in predicting behaviors in contained environments like the auditorium.