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Chapter 14: Problem 27

Imagine that 501 people are present in a movie theater of volume $8.00 \times 10^{3} \mathrm{m}^{3}$ that is sealed shut so no air can escape. Each person gives off heat at an average rate of \(110 \mathrm{W} .\) By how much will the temperature of the air have increased during a 2.0 -h movie? The initial pressure is \(1.01 \times 10^{5} \mathrm{Pa}\) and the initial temperature is \(20.0^{\circ} \mathrm{C} .\) Assume that all the heat output of the people goes into heating the air (a diatomic gas).

Short Answer

Expert verified
Based on the given information and calculations, the increase in temperature of the air inside the sealed movie theater after the 2-hour duration of the movie is approximately 9.52 K.

Step by step solution

01

Find the total heat output of the people

First, we need to find the total heat output from all the people after the 2-hour duration of a movie. Multiply the average heat output per person by the number of people and the duration of the movie, which we must convert to seconds: Total heat output = (Number of people) × (average heat output per person) × (time in seconds) Total heat output = 501 × 110 W × (2.0 hours × 3600 s/hour) = 3.32 × 10^7 J
02

Use the ideal gas law to find the initial number of moles in the theater

We need to find the number of moles of air inside the theater initially so that we can calculate the change in temperature due to the heat added. To do this, we will use the ideal gas law, as follows: PV = nRT Here, P = Initial pressure = 1.01 × 10^5 Pa V = Volume of theater = 8.00 × 10^3 m³ R = Ideal gas constant = 8.314 J/(mol K) T = Initial temperature, which we must convert to Kelvin: 20.0°C + 273.15 = 293.15 K Now, solve for the number of moles, n: n = PV/(RT) = (1.01 × 10⁵ Pa)(8.00 × 10³ m³) / (8.314 J/mol K)(293.15 K) = 3.50 × 10⁴ mol
03

Calculate the increase in temperature using the heat capacity of diatomic gases

Now that we know the number of moles and the total heat output, we can use the heat capacity of diatomic gases to calculate the change in temperature. The molar heat capacity at constant volume, C_v, for diatomic gases (air can be considered a diatomic gas) is: C_v = (5/2)R, where R = 8.314 J/(mol K) The total heat added to the system can be expressed as: ΔQ = n × C_v × ΔT Rearrange and solve for the change in temperature, ΔT: ΔT = ΔQ / (n × C_v) = (3.32 × 10⁷ J) / (3.50 × 10⁴ mol × (5/2)(8.314 J/mol K)) = 9.52 K Since the question asks for the increase in temperature, our final answer is: Temperature increase = 9.52 K

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Most popular questions from this chapter

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