2) Two bottles at the same temperature contain the same kind of gas. Bottle B has twice the volume and half the mass of gas as Bottle A. How does the pressure in B compare with the pressure in A? A) PB = % PA В) Рв 3 2 РА C) PB = 4 PA D) PB = 4 PA E) PB = PA
2) Two bottles at the same temperature contain the same kind of gas. Bottle B has twice the volume and half the mass of gas as Bottle A. How does the pressure in B compare with the pressure in A? A) PB = % PA В) Рв 3 2 РА C) PB = 4 PA D) PB = 4 PA E) PB = PA
Related questions
Question
![### Gas Pressure Comparison
**Problem Statement:**
Two bottles at the same temperature contain the same kind of gas. Bottle B has twice the volume and half the mass of gas as Bottle A. How does the pressure in Bottle B (PB) compare with the pressure in Bottle A (PA)?
**Options:**
A) \( P_B = \frac{1}{2} P_A \)
B) \( P_B = 2 P_A \)
C) \( P_B = \frac{1}{4} P_A \)
D) \( P_B = 4 P_A \)
E) \( P_B = P_A \)
**Explanation:**
To solve this, we can use the Ideal Gas Law, which states:
\[
PV = nRT
\]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. Since the bottles contain the same gas at the same temperature, \( T \) and \( R \) are constant.
Given:
1. Volume of Bottle B (\( V_B \)) is twice that of Bottle A (\( V_A \)).
\[
V_B = 2V_A
\]
2. Mass of gas in Bottle B is half that in Bottle A. Since the gas is the same, this also means the number of moles in Bottle B (\( n_B \)) is half that in Bottle A (\( n_A \)).
\[
n_B = \frac{1}{2} n_A
\]
Using the Ideal Gas Law for both bottles:
For Bottle A:
\[
P_A V_A = n_A RT
\]
For Bottle B:
\[
P_B V_B = n_B RT
\]
Substitute the given relationships into the equations:
\[
P_B \cdot 2V_A = \left( \frac{1}{2} n_A \right) RT
\]
Simplify the equation:
\[
P_B \cdot 2V_A = \frac{1}{2} n_A RT
\]
\[
P_B \cdot 2V_A = \frac{1}{2} P_A V_A
\]
\[
P_B = \frac{1}{4} P_A
\]
Thus,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5aab90fe-f662-4329-be2f-bc153bcfca52%2F3c923aae-ca84-45c4-be0f-0aef9c737b7d%2F3mztyzo_processed.png&w=3840&q=75)
Transcribed Image Text:### Gas Pressure Comparison
**Problem Statement:**
Two bottles at the same temperature contain the same kind of gas. Bottle B has twice the volume and half the mass of gas as Bottle A. How does the pressure in Bottle B (PB) compare with the pressure in Bottle A (PA)?
**Options:**
A) \( P_B = \frac{1}{2} P_A \)
B) \( P_B = 2 P_A \)
C) \( P_B = \frac{1}{4} P_A \)
D) \( P_B = 4 P_A \)
E) \( P_B = P_A \)
**Explanation:**
To solve this, we can use the Ideal Gas Law, which states:
\[
PV = nRT
\]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. Since the bottles contain the same gas at the same temperature, \( T \) and \( R \) are constant.
Given:
1. Volume of Bottle B (\( V_B \)) is twice that of Bottle A (\( V_A \)).
\[
V_B = 2V_A
\]
2. Mass of gas in Bottle B is half that in Bottle A. Since the gas is the same, this also means the number of moles in Bottle B (\( n_B \)) is half that in Bottle A (\( n_A \)).
\[
n_B = \frac{1}{2} n_A
\]
Using the Ideal Gas Law for both bottles:
For Bottle A:
\[
P_A V_A = n_A RT
\]
For Bottle B:
\[
P_B V_B = n_B RT
\]
Substitute the given relationships into the equations:
\[
P_B \cdot 2V_A = \left( \frac{1}{2} n_A \right) RT
\]
Simplify the equation:
\[
P_B \cdot 2V_A = \frac{1}{2} n_A RT
\]
\[
P_B \cdot 2V_A = \frac{1}{2} P_A V_A
\]
\[
P_B = \frac{1}{4} P_A
\]
Thus,
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
