Use Gay-Lussac equation to solve the given problem." 1. A certain light bulb containing argon has a pressure 1.20 atm at 18 0 C. If it will be heated to 850C at constant volume, what will be the resulting pressure ( P2 )? Is it enough to cause sudden breakage of the bulb?
Use Gay-Lussac equation to solve the given problem." 1. A certain light bulb containing argon has a pressure 1.20 atm at 18 0 C. If it will be heated to 850C at constant volume, what will be the resulting pressure ( P2 )? Is it enough to cause sudden breakage of the bulb?
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Use Gay-Lussac equation to solve the given problem."
1. A certain light bulb containing argon has a pressure 1.20 atm at 18 0 C. If it will be heated to 850C at
constant volume, what will be the resulting pressure ( P2 )? Is it enough to cause sudden breakage of the
bulb?
2. At 293 K a confined ammonia gas has a pressure of 2.50 atm. At what temperature ( T2 ) would its
pressure be equal to 760 mmHg ?
![**Gay-Lussac's Law**
Gay-Lussac's Law states that at constant volume, the pressure of a fixed mass of gas is directly proportional to the absolute temperature. In other words, if the temperature increases or decreases, the pressure will also increase or decrease respectively.
The mathematical representation of Gay-Lussac's Law is:
\[ P = kT \]
Or, in terms of relations between two states:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Alternatively, it can be written as:
\[ P_1 T_2 = P_2 T_1 \]
### Sample Problem:
**Problem Statement:**
A helium tank has a pressure of 650 torr at 25°C. What will be the pressure (P₂) if the temperature is tripled?
**Given:**
- Initial pressure, \( P_1 \) = 650 torr
- Initial temperature, \( T_1 \) = 25°C = 25 + 273 = 298 K
- Final temperature, \( T_2 \) = 3 × (25°C) + 273 = 348 K
**Unknown:**
- Final pressure \( P_2 = ? \)
**Formula:**
\[ P_2 = P_1 \frac{T_2}{T_1} \]
**Solution:**
\[ P_2 = (650 \text{ torr}) \frac{348 \text{ K}}{298 \text{ K}} \]
**Answer:**
\[ P_2 = 759 \text{ torr} \]
The image provided includes this explanation and lays out the steps to solve the problem using the relevant formula. This illustration helps in understanding how changes in temperature affect the pressure of a gas at constant volume according to Gay-Lussac's Law.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61c4f236-74fc-421f-b5e2-11da5c6f14f1%2F5dd0685d-5b7f-4fc3-86f8-047b0c23386c%2F6cq78ie_processed.png&w=3840&q=75)
Transcribed Image Text:**Gay-Lussac's Law**
Gay-Lussac's Law states that at constant volume, the pressure of a fixed mass of gas is directly proportional to the absolute temperature. In other words, if the temperature increases or decreases, the pressure will also increase or decrease respectively.
The mathematical representation of Gay-Lussac's Law is:
\[ P = kT \]
Or, in terms of relations between two states:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Alternatively, it can be written as:
\[ P_1 T_2 = P_2 T_1 \]
### Sample Problem:
**Problem Statement:**
A helium tank has a pressure of 650 torr at 25°C. What will be the pressure (P₂) if the temperature is tripled?
**Given:**
- Initial pressure, \( P_1 \) = 650 torr
- Initial temperature, \( T_1 \) = 25°C = 25 + 273 = 298 K
- Final temperature, \( T_2 \) = 3 × (25°C) + 273 = 348 K
**Unknown:**
- Final pressure \( P_2 = ? \)
**Formula:**
\[ P_2 = P_1 \frac{T_2}{T_1} \]
**Solution:**
\[ P_2 = (650 \text{ torr}) \frac{348 \text{ K}}{298 \text{ K}} \]
**Answer:**
\[ P_2 = 759 \text{ torr} \]
The image provided includes this explanation and lays out the steps to solve the problem using the relevant formula. This illustration helps in understanding how changes in temperature affect the pressure of a gas at constant volume according to Gay-Lussac's Law.
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