### Air Expansion in Food Packaging Due to Elevation Changes1. **Observing Changes in Packaged Food at High Elevation:** People buying food in sealed bags at high elevation often notice that the bags are puffed up because the air inside has expanded. Consider a scenario where: - A bag of pretzels was initially packed at a sea-level pressure of 1.00 atm and a temperature of 22.0 °C. - At a summer picnic in Santa Fe, New Mexico, with an ambient temperature of 32.0 °C, the volume of air in the bag is observed to be 1.19 times its original volume. The question posed is: What is the pressure of the air \( P_{\text{Santa Fe}} \)? This can be calculated using the ideal gas law principle, adjusting for temperature and volume changes, recognizing that temperature must be in kelvin for accurate calculation.2. **Calculation Section:** - Convert temperatures from Celsius to Kelvin: - Initial temperature \( T_1 \): \( 22.0 \,°C = 295.15 \, K \) - Final temperature \( T_2 \): \( 32.0 \,°C = 305.15 \, K \) - Apply the combined gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) - Where: - \( P_1 = 1.00 \, atm \) - \( V_1 = 1 \) (initial volume) - \( V_2 = 1.19 \, V_1 \) (final volume) - \( T_1 = 295.15 \, K \) - \( T_2 = 305.15 \, K \) - Rearrange and solve for \( P_2 \) (Pressure in Santa Fe): \[ P_{\text{Santa Fe}} = \frac{P_1 \cdot V_1 \cdot T_2}{V_2 \cdot T_1} = \frac{1.00 \cdot 1 \cdot 305.15}{1.19 \cdot 295.15} \] \[ P_{\text{Santa Fe}} \approx 0

We will consider air inside the bag as ideal gas, and use ideal gas equation to solve the problem.…

Answered: 1. People buying food in sealed bags at…

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