1. People buying food in sealed bags at high elevation often notice that the bags are puffed up because the air inside has expanded. A bag of pretzels was packed at a pressure of 1.00 atm and a temperature of 22.0 °C. When opened at a summer picnic in Santa Fe, New Mexico, at a temperature of 32.0 °C, the volume of air in the bag is 1.33 times its original volume. What is the pressure of the Santa Fe? air P P Santa Fe = atm

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### Real-World Applications of Gas Laws: High Elevation and Sealed Food Bags

#### Understanding Pressure Changes in Sealed Bags

When people purchase food in sealed bags at high elevations, they often observe that the bags are puffed up. This phenomenon occurs because the air inside the bags expands due to changes in atmospheric conditions.

Consider the following example to understand this better:

A bag of pretzels was packed at sea level with the following conditions:
- **Pressure (P₁):** 1.00 atm
- **Temperature (T₁):** 22.0°C

When the bag is later opened at a summer picnic in Santa Fe, New Mexico, new conditions are observed:
- **Temperature (T₂):** 32.0°C
- **Volume (V₂):** 1.33 times the original volume (V₁)

### Question:
What is the pressure of the air (\( P_{\text{Santa Fe}} \)) inside the bag after it expands?

### Formula:
To find the final pressure (\( P_{\text{Santa Fe}} \)), we can use the combined gas law:

\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]

Given parameters:
- **P₁ = 1.00 atm**
- **V₁ = Initial volume**
- **T₁ = 22.0°C** (converted to Kelvin: 22.0 + 273.15 = 295.15 K)
- **V₂ = 1.33 V₁**
- **T₂ = 32.0°C** (converted to Kelvin: 32.0 + 273.15 = 305.15 K)
- **P₂ = \( P_{\text{Santa Fe}} \)**

Substitute these values into the equation, solving for \( P_{\text{Santa Fe}} \):

\[ \frac{1.00 \, \text{atm} \times V₁}{295.15 \, \text{K}} = \frac{P_{\text{Santa Fe}} \times 1.33 V₁}{305.15 \, \text{K}} \]

By cancelling \( V₁ \) from both sides and solving for \( P_{\text{Santa Fe}} \):

\[ P_{\text{Santa
Transcribed Image Text:### Real-World Applications of Gas Laws: High Elevation and Sealed Food Bags #### Understanding Pressure Changes in Sealed Bags When people purchase food in sealed bags at high elevations, they often observe that the bags are puffed up. This phenomenon occurs because the air inside the bags expands due to changes in atmospheric conditions. Consider the following example to understand this better: A bag of pretzels was packed at sea level with the following conditions: - **Pressure (P₁):** 1.00 atm - **Temperature (T₁):** 22.0°C When the bag is later opened at a summer picnic in Santa Fe, New Mexico, new conditions are observed: - **Temperature (T₂):** 32.0°C - **Volume (V₂):** 1.33 times the original volume (V₁) ### Question: What is the pressure of the air (\( P_{\text{Santa Fe}} \)) inside the bag after it expands? ### Formula: To find the final pressure (\( P_{\text{Santa Fe}} \)), we can use the combined gas law: \[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \] Given parameters: - **P₁ = 1.00 atm** - **V₁ = Initial volume** - **T₁ = 22.0°C** (converted to Kelvin: 22.0 + 273.15 = 295.15 K) - **V₂ = 1.33 V₁** - **T₂ = 32.0°C** (converted to Kelvin: 32.0 + 273.15 = 305.15 K) - **P₂ = \( P_{\text{Santa Fe}} \)** Substitute these values into the equation, solving for \( P_{\text{Santa Fe}} \): \[ \frac{1.00 \, \text{atm} \times V₁}{295.15 \, \text{K}} = \frac{P_{\text{Santa Fe}} \times 1.33 V₁}{305.15 \, \text{K}} \] By cancelling \( V₁ \) from both sides and solving for \( P_{\text{Santa Fe}} \): \[ P_{\text{Santa
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