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 0.770 atm, then the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?
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 0.770 atm, then the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?
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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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Temperature chemistry homework problem please
![**Title: Calculating Temperature Change at Elevated Atmospheric Pressure**
**Problem Statement:**
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 0.770 atm, the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?
**Solution:**
To solve this problem, we can use the combined gas law. The combined gas law states that for a given amount of gas, the ratio of the product of pressure and volume to the temperature remains constant:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures
- \( V_1 \) and \( V_2 \) are the initial and final volumes
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin
First, convert the initial temperature from Celsius to Kelvin:
\[ T_1 = 25.0°C + 273.15 = 298.15 K \]
Given values:
- \( P_1 = 1.00 \, \text{atm} \)
- \( V_1 = 1.20 \, \text{L} \)
- \( P_2 = 0.770 \, \text{atm} \)
- \( V_2 = 1.80 \, \text{L} \)
We need to find \( T_2 \) in Kelvin and then convert it to Celsius. Rearrange the combined gas law to solve for \( T_2 \):
\[ T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \]
Substitute the known values into the equation:
\[ T_2 = \frac{(0.770 \, \text{atm}) (1.80 \, \text{L}) (298.15 \, K)}{(1.00 \, \text{atm}) (1.20 \, \text{L})} \]
\[ T_2 = \frac{0.770 \times 1.80 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ae9165-8c64-40e9-9a68-65bc31ca98b7%2Fc0f3e99c-81be-4f69-bcc6-fffb183b0352%2Ftgsvib_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Calculating Temperature Change at Elevated Atmospheric Pressure**
**Problem Statement:**
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 0.770 atm, the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?
**Solution:**
To solve this problem, we can use the combined gas law. The combined gas law states that for a given amount of gas, the ratio of the product of pressure and volume to the temperature remains constant:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures
- \( V_1 \) and \( V_2 \) are the initial and final volumes
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin
First, convert the initial temperature from Celsius to Kelvin:
\[ T_1 = 25.0°C + 273.15 = 298.15 K \]
Given values:
- \( P_1 = 1.00 \, \text{atm} \)
- \( V_1 = 1.20 \, \text{L} \)
- \( P_2 = 0.770 \, \text{atm} \)
- \( V_2 = 1.80 \, \text{L} \)
We need to find \( T_2 \) in Kelvin and then convert it to Celsius. Rearrange the combined gas law to solve for \( T_2 \):
\[ T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \]
Substitute the known values into the equation:
\[ T_2 = \frac{(0.770 \, \text{atm}) (1.80 \, \text{L}) (298.15 \, K)}{(1.00 \, \text{atm}) (1.20 \, \text{L})} \]
\[ T_2 = \frac{0.770 \times 1.80 \times
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