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A truck tire is filled with air to an interior pressure of 3.11 atm at 15.1

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C. When the truck drives down the highway, the temperature of the tire and air inside increases. Assume that the tire is in good working condition so that no air escapes as the temperature changes. After driving for a while, the volume of the tire has increased by a factor of 0.025( V

f

​

=(1+0.025)V

i

​

). The temperature has increased to 37.9

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C. What is the final pressure of the air in the tire? Give your answer in atm to at least three significant digits. NOTE: For those of you who know the difference, the pressures here are the absolute pressures of the air in the tire, NOT the gauge pressure. For those of you who don't know the different, the "gauge pressure" is (not surprisingly) what most pressure gauges measure, which is the DIFFERENCE between the pressure inside a container and the pressure outside the container. So, the absolute pressure, in general, is the gauge pressure plus one atm. (You DON'T need to worry about this, because the pressures given are the absolute pressures.)

Answer :

The final pressure of the air in the tire is 3.18 atm.

When the temperature of the tire and air inside increases, the volume of the tire also increases. In this case, the volume of the tire has increased by a factor of 0.025. To determine the final pressure, we can apply the ideal gas law, which states that the product of pressure and volume is directly proportional to the temperature.

Using the formula: P₁V₁/T₁ = P₂V₂/T₂, where P₁, V₁, and T₁ are the initial pressure, volume, and temperature, and P₂, V₂, and T₂ are the final pressure, volume, and temperature, respectively, we can solve for the final pressure.

Given:

P₁ = 3.11 atm (initial pressure)

V₁ = V (initial volume)

T₁ = 15.1 °C = 15.1 + 273.15 = 288.25 K (initial temperature)

V₂ = (1 + 0.025)V (final volume)

T₂ = 37.9 °C = 37.9 + 273.15 = 311.05 K (final temperature)

Applying the values to the formula and solving for P₂:

3.11 atm × V / 288.25 K = P₂ × (1 + 0.025)V / 311.05 K

P₂ = (3.11 atm × 311.05 K) / (288.25 K × (1 + 0.025))

P₂ ≈ 3.18 atm

Therefore, the final pressure of the air in the tire is approximately 3.18 atm.

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When a truck tire drives down the highway, the increase in temperature leads to an expansion of the tire and the air inside it. This increase in volume can be quantified by a factor of 0.025.

To determine the final pressure of the air, the ideal gas law is utilized, which establishes a relationship between pressure, volume, and temperature. By applying the ideal gas law equation (P₁V₁/T₁ = P₂V₂/T₂) and plugging in the given values, we can solve for the final pressure.

In this case, the initial pressure (P₁) is 3.11 atm, the initial volume (V₁) is denoted as V, and the initial temperature (T₁) is 15.1 °C (or 288.25 K). The final volume (V₂) is calculated as (1 + 0.025)V, and the final temperature (T₂) is 37.9 °C (or 311.05 K). Substituting these values into the ideal gas law equation allows us to find the final pressure (P₂).

After solving the equation, we determine that the final pressure of the air in the tire is approximately 3.18 atm. This means that as the tire heats up and expands, the pressure within the tire also increases.

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