Suppose 554 mL of argon at 785 torr is cooled from 25.0 °C to -134.0 °C. What will be the volume of the sample at the lower temperature and 1.45 atm?
Suppose 554 mL of argon at 785 torr is cooled from 25.0 °C to -134.0 °C. What will be the volume of the sample at the lower temperature and 1.45 atm?
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter5: The Gaseous State
Section: Chapter Questions
Problem 5.28QP
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Question
![**Problem Statement:**
Suppose 554 mL of argon at 785 torr is cooled from 25.0°C to -134.0°C. What will be the volume of the sample at the lower temperature and 1.45 atm?
---
**Analysis:**
To solve this problem, we will use the Combined Gas Law which is derived from Boyle's Law, Charles's Law, and Gay-Lussac's Law. The Combined Gas Law formula is:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
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, we need to convert all units to the appropriate SI units:
1. **Convert pressure from torr to atm:**
- Given \( P_1 = 785 \text{ torr} \)
- Conversion: \( 1 \text{ atm} = 760 \text{ torr} \), hence \( P_1 = \frac{785}{760} \text{ atm} \approx 1.033 \text{ atm} \)
2. **Convert temperatures from degrees Celsius to Kelvin:**
- \( T_1 = 25.0 \degree C = 25.0 + 273.15 = 298.15 \text{ K} \)
- \( T_2 = -134.0 \degree C = -134.0 + 273.15 = 139.15 \text{ K} \)
3. **Utilize the given values and plug them into the Combined Gas Law:**
\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\]
\[
\frac{1.033 \text{ atm} \times 554 \text{ mL}}{298.15 \text{ K}} = \frac{1.45 \text{ atm} \times V_2}{139.15 \text{ K}}
\]
4. **Solve for \( V_2 \):**
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f4e1179-0845-4cce-a522-94afec860b33%2Fe44d8de9-9d5c-4297-89ff-d8e5d941f829%2Fn81pw6p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose 554 mL of argon at 785 torr is cooled from 25.0°C to -134.0°C. What will be the volume of the sample at the lower temperature and 1.45 atm?
---
**Analysis:**
To solve this problem, we will use the Combined Gas Law which is derived from Boyle's Law, Charles's Law, and Gay-Lussac's Law. The Combined Gas Law formula is:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
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, we need to convert all units to the appropriate SI units:
1. **Convert pressure from torr to atm:**
- Given \( P_1 = 785 \text{ torr} \)
- Conversion: \( 1 \text{ atm} = 760 \text{ torr} \), hence \( P_1 = \frac{785}{760} \text{ atm} \approx 1.033 \text{ atm} \)
2. **Convert temperatures from degrees Celsius to Kelvin:**
- \( T_1 = 25.0 \degree C = 25.0 + 273.15 = 298.15 \text{ K} \)
- \( T_2 = -134.0 \degree C = -134.0 + 273.15 = 139.15 \text{ K} \)
3. **Utilize the given values and plug them into the Combined Gas Law:**
\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\]
\[
\frac{1.033 \text{ atm} \times 554 \text{ mL}}{298.15 \text{ K}} = \frac{1.45 \text{ atm} \times V_2}{139.15 \text{ K}}
\]
4. **Solve for \( V_2 \):**
\[
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