< Question 20 of 46 A 1.20 L weather balloon on the ground has a temperature of 25.0 °C and is at atmospheric pressure (1.00 atm). When it rises to an elevation where the pressure is 0.750 atm, then the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?
< Question 20 of 46 A 1.20 L weather balloon on the ground has a temperature of 25.0 °C and is at atmospheric pressure (1.00 atm). When it rises to an elevation where the pressure is 0.750 atm, then the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?
Chemistry
10th Edition
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Title**: Weather Balloon Temperature Calculation
**Question**:
A 12.0 L weather balloon on the ground has a temperature of 25.0°C and is at atmospheric pressure (1.00 atm). When it rises to an elevation where the pressure is 0.750 atm, then the volume is 18.0 L. What is the temperature (in °C) of the air at this elevation?
**Answer Input**:
- Temperature (°C): (Input field with number pad)
[Submit Button]
**Explanation**:
To solve this problem, use the Combined Gas Law which relates pressure (P), volume (V), and temperature (T):
\[
\frac{{P_1 \cdot V_1}}{{T_1}} = \frac{{P_2 \cdot V_2}}{{T_2}}
\]
Where:
- \( P_1 \) = Initial pressure = 1.00 atm
- \( V_1 \) = Initial volume = 12.0 L
- \( T_1 \) = Initial temperature = 25.0°C (converted to Kelvin = 298.15 K) (*Remember: Kelvin = °C + 273.15*)
- \( P_2 \) = Final pressure = 0.750 atm
- \( V_2 \) = Final volume = 18.0 L
- \( T_2 \) = Final temperature in Kelvin
Step-by-step solution:
1. Convert the initial temperature from Celsius to Kelvin:
\[
T_1 = 25.0 + 273.15 = 298.15 \text{ K}
\]
2. Substitute the known values into the Combined Gas Law:
\[
\frac{{1.00 \text{ atm} \cdot 12.0 \text{ L}}}{{298.15 \text{ K}}} = \frac{{0.750 \text{ atm} \cdot 18.0 \text{ L}}}{{T_2}}
\]
3. Solve for \( T_2 \):
\[
T_2 = \frac{{0.750 \cdot 18.0 \cdot 298.15}}{{1.00 \cdot 12.0}} = 336.67 \text{ K}
\]
4. Convert the final temperature back to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20214f54-b261-4746-a2af-8e52bbd8e109%2F58ffc7d3-c1af-4f66-8472-fc971fdff507%2Fdwmonxk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
**Title**: Weather Balloon Temperature Calculation
**Question**:
A 12.0 L weather balloon on the ground has a temperature of 25.0°C and is at atmospheric pressure (1.00 atm). When it rises to an elevation where the pressure is 0.750 atm, then the volume is 18.0 L. What is the temperature (in °C) of the air at this elevation?
**Answer Input**:
- Temperature (°C): (Input field with number pad)
[Submit Button]
**Explanation**:
To solve this problem, use the Combined Gas Law which relates pressure (P), volume (V), and temperature (T):
\[
\frac{{P_1 \cdot V_1}}{{T_1}} = \frac{{P_2 \cdot V_2}}{{T_2}}
\]
Where:
- \( P_1 \) = Initial pressure = 1.00 atm
- \( V_1 \) = Initial volume = 12.0 L
- \( T_1 \) = Initial temperature = 25.0°C (converted to Kelvin = 298.15 K) (*Remember: Kelvin = °C + 273.15*)
- \( P_2 \) = Final pressure = 0.750 atm
- \( V_2 \) = Final volume = 18.0 L
- \( T_2 \) = Final temperature in Kelvin
Step-by-step solution:
1. Convert the initial temperature from Celsius to Kelvin:
\[
T_1 = 25.0 + 273.15 = 298.15 \text{ K}
\]
2. Substitute the known values into the Combined Gas Law:
\[
\frac{{1.00 \text{ atm} \cdot 12.0 \text{ L}}}{{298.15 \text{ K}}} = \frac{{0.750 \text{ atm} \cdot 18.0 \text{ L}}}{{T_2}}
\]
3. Solve for \( T_2 \):
\[
T_2 = \frac{{0.750 \cdot 18.0 \cdot 298.15}}{{1.00 \cdot 12.0}} = 336.67 \text{ K}
\]
4. Convert the final temperature back to
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