A sample of hydrogen occupies 331.0 mL at STP. If the temperature were increased to 518.0°C, what final pressure (atm) would be necessary to keep the volume constant at 331.0 mi Submit Answer Tries 0/99 This discussion is closed. Send R
A sample of hydrogen occupies 331.0 mL at STP. If the temperature were increased to 518.0°C, what final pressure (atm) would be necessary to keep the volume constant at 331.0 mi Submit Answer Tries 0/99 This discussion is closed. Send R
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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![### Educational Resource: Gas Law Calculations
#### Problem Statement:
A sample of hydrogen occupies 331.0 mL at STP (Standard Temperature and Pressure). If the temperature were increased to 518.0°C, what final pressure (atm) would be necessary to keep the volume constant at 331.0 mL?
#### Discussion Status:
- **This discussion is closed.**
### Explanation:
When dealing with gases and changes in temperature and pressure, one of the fundamental principles used is the Combined Gas Law. This law relates the pressure, volume, and temperature of a gas, allowing us to determine unknown variables in different conditions when certain parameters are held constant.
The formula for the Combined Gas Law is:
\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]
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.
In this problem:
- \( V_1 = V_2 = 331.0 \) mL (volume remains constant)
- Initial temperature, \( T_1 \), is 273.15 K (STP)
- Final temperature, \( T_2 \), is 791.15 K (converted from 518.0°C)
- We need to find \( P_2 \) while \( P_1 \) is 1 atm at STP.
Since the volume is constant, it simplifies to a direct relationship between pressure and temperature:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
This can be rearranged to solve for \( P_2 \):
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]
### Conclusion:
By substituting the known values, we can calculate the final pressure required to keep the volume constant with the increased temperature.
(Note: For the actual numerical solution, you would need to compute based on the provided values, ensuring that temperatures are converted to Kelvin.)
### Additional Note:
Software tools, typically like the one shown in the image, often limit the number of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d7f79bc-69ba-4efa-950e-7604f8442ee4%2F4c8041f2-8461-4cf0-8644-f70a5ba12e93%2F8nz3g4f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Resource: Gas Law Calculations
#### Problem Statement:
A sample of hydrogen occupies 331.0 mL at STP (Standard Temperature and Pressure). If the temperature were increased to 518.0°C, what final pressure (atm) would be necessary to keep the volume constant at 331.0 mL?
#### Discussion Status:
- **This discussion is closed.**
### Explanation:
When dealing with gases and changes in temperature and pressure, one of the fundamental principles used is the Combined Gas Law. This law relates the pressure, volume, and temperature of a gas, allowing us to determine unknown variables in different conditions when certain parameters are held constant.
The formula for the Combined Gas Law is:
\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]
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.
In this problem:
- \( V_1 = V_2 = 331.0 \) mL (volume remains constant)
- Initial temperature, \( T_1 \), is 273.15 K (STP)
- Final temperature, \( T_2 \), is 791.15 K (converted from 518.0°C)
- We need to find \( P_2 \) while \( P_1 \) is 1 atm at STP.
Since the volume is constant, it simplifies to a direct relationship between pressure and temperature:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
This can be rearranged to solve for \( P_2 \):
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]
### Conclusion:
By substituting the known values, we can calculate the final pressure required to keep the volume constant with the increased temperature.
(Note: For the actual numerical solution, you would need to compute based on the provided values, ensuring that temperatures are converted to Kelvin.)
### Additional Note:
Software tools, typically like the one shown in the image, often limit the number of
Expert Solution
Step 1
Given data;
Initial volume oh Hydrogen = 331 mL
Initial temperature = 273.15 K (Given that the initial condition is STP)
Initial pressure = 1 atm ( according to STP)
Final temperature = 5180C
Inorder to convert degree celsius to kelvin we need to add 273.15 to the degree celsius value. Thus
5180C = 518+ 273.15 = 791.15 K
It is mentioned that volume has to be remained as 331 mL, that is it needs to be kept constant. And we have to find the final pressure to maintain these conditions.
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