A gas in a can of spray paint is initially at room temperature (30°C). On a particularly hot day, the temperature of the gas in the can reaches 63°C and exerts a pressure of 4 atm. What was the initial pressure of the gas in the can of spray paint, in atmospheres?
A gas in a can of spray paint is initially at room temperature (30°C). On a particularly hot day, the temperature of the gas in the can reaches 63°C and exerts a pressure of 4 atm. What was the initial pressure of the gas in the can of spray paint, in atmospheres?
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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![**Problem Statement:**
A gas in a can of spray paint is initially at room temperature (30°C). On a particularly hot day, the temperature of the gas in the can reaches 63°C and exerts a pressure of 4 atm. What was the initial pressure of the gas in the can of spray paint, in atmospheres?
**Explanation:**
To solve this problem, we'll use the ideal gas law in its simpler form that relates pressure and temperature, since the volume and amount of gas remain constant. This is known as Gay-Lussac's law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \( P_1 \) is the initial pressure.
- \( T_1 \) is the initial temperature in Kelvin.
- \( P_2 \) is the final pressure.
- \( T_2 \) is the final temperature in Kelvin.
**Steps to Solve:**
1. Convert the temperatures from Celsius to Kelvin:
\[
T_1 = 30°C + 273.15 = 303.15 \, K
\]
\[
T_2 = 63°C + 273.15 = 336.15 \, K
\]
2. Use the formula to find the initial pressure \( P_1 \):
\[
\frac{P_1}{303.15} = \frac{4}{336.15}
\]
3. Solve for \( P_1 \):
\[
P_1 = \frac{4 \times 303.15}{336.15} = 3.61 \, \text{atm}
\]
The initial pressure of the gas in the can is approximately 3.61 atm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9dbaf28b-531f-4842-8caf-eb3ad75d7a35%2F647f1442-f4e1-418b-a36b-36eeae4976d9%2Foipmqm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A gas in a can of spray paint is initially at room temperature (30°C). On a particularly hot day, the temperature of the gas in the can reaches 63°C and exerts a pressure of 4 atm. What was the initial pressure of the gas in the can of spray paint, in atmospheres?
**Explanation:**
To solve this problem, we'll use the ideal gas law in its simpler form that relates pressure and temperature, since the volume and amount of gas remain constant. This is known as Gay-Lussac's law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \( P_1 \) is the initial pressure.
- \( T_1 \) is the initial temperature in Kelvin.
- \( P_2 \) is the final pressure.
- \( T_2 \) is the final temperature in Kelvin.
**Steps to Solve:**
1. Convert the temperatures from Celsius to Kelvin:
\[
T_1 = 30°C + 273.15 = 303.15 \, K
\]
\[
T_2 = 63°C + 273.15 = 336.15 \, K
\]
2. Use the formula to find the initial pressure \( P_1 \):
\[
\frac{P_1}{303.15} = \frac{4}{336.15}
\]
3. Solve for \( P_1 \):
\[
P_1 = \frac{4 \times 303.15}{336.15} = 3.61 \, \text{atm}
\]
The initial pressure of the gas in the can is approximately 3.61 atm.
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