any drivers report low tire pressure at the start of winter when the temperature drops. If the pressure in a tire is 2.38 atm when the outside temperature is 294 K, what will the essure in the tire be if the outside temperature drops to 280. K? (Assume that the volume and atmospheric pressure remain constant.) A B 0.170 atm 2.27 atm

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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### Understanding Tire Pressure Changes with Temperature

**Scenario:**

Many drivers report low tire pressure at the start of winter when the temperature drops. Consider the following situation: 

- The pressure in a tire is 2.38 atm when the outside temperature is 294 K.
- What will the pressure in the tire be if the outside temperature drops to 280 K?

**Assumptions:**

- The volume and atmospheric pressure remain constant.

**Options:**

- **A**: 0.170 atm
- **B**: 2.27 atm
- **C**: 2.50 atm
- **D**: 14.0 atm

---

To solve problems like this, we use the ideal gas law in the form of the combined gas law, which relates pressure and temperature when volume is constant: 

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures
- \( T_1 \) and \( T_2 \) are the initial and final temperatures (in Kelvin)

By plugging in the values:
- \( P_1 = 2.38 \) atm
- \( T_1 = 294 \) K
- \( T_2 = 280 \) K

You can solve for \( P_2 \) to find the final pressure in the tire.
Transcribed Image Text:### Understanding Tire Pressure Changes with Temperature **Scenario:** Many drivers report low tire pressure at the start of winter when the temperature drops. Consider the following situation: - The pressure in a tire is 2.38 atm when the outside temperature is 294 K. - What will the pressure in the tire be if the outside temperature drops to 280 K? **Assumptions:** - The volume and atmospheric pressure remain constant. **Options:** - **A**: 0.170 atm - **B**: 2.27 atm - **C**: 2.50 atm - **D**: 14.0 atm --- To solve problems like this, we use the ideal gas law in the form of the combined gas law, which relates pressure and temperature when volume is constant: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Where: - \( P_1 \) and \( P_2 \) are the initial and final pressures - \( T_1 \) and \( T_2 \) are the initial and final temperatures (in Kelvin) By plugging in the values: - \( P_1 = 2.38 \) atm - \( T_1 = 294 \) K - \( T_2 = 280 \) K You can solve for \( P_2 \) to find the final pressure in the tire.
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