In a refrigerator, 2.10 mol of an ideal monatomic gas is taken through the cycle shown in the figure. The temperature at point A is 782.0 K What is the temperature at point D? K
In a refrigerator, 2.10 mol of an ideal monatomic gas is taken through the cycle shown in the figure. The temperature at point A is 782.0 K What is the temperature at point D? K
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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In a refrigerator, 2.10 mol of an ideal monatomic gas is taken through the cycle shown in the figure. The temperature at point A is 782.0 K
What is the temperature at point D? K
![### Ideal Gas Cycle in Refrigerators: Temperature Calculation
In this example, we will explore the thermodynamic cycle of an ideal monatomic gas within a refrigerator. The gas amount given is 2.10 moles, and we are asked to determine the temperature at point \( D \) on the presented cycle diagram. It is given that the temperature at point \( A \) is 782.0 K.
#### Diagram Analysis
The diagram displays a Pressure-Volume (P-V) cycle of the gas. The cycle consists of four points labeled A, B, C, and D, and it can be broken down into a series of processes:
- From \( A \) to \( D \): Isobaric (constant pressure) expansion.
- From \( D \) to \( C \): Isochoric (constant volume) cooling.
- From \( C \) to \( B \): Isobaric (constant pressure) compression.
- From \( B \) to \( A \): Isochoric (constant volume) heating.
The key values depicted in the diagram include:
- Pressure \( P_2 \) at points \( A \) and \( D \).
- Pressure \( 1.30 \text{ kPa} \) at points \( B \) and \( C \).
- Volume \( 1.50 \text{ m}^3 \) at points \( A \) and \( B \).
- Volume \( 2.25 \text{ m}^3 \) at points \( C \) and \( D \).
#### Given
- Moles of gas (\( n \)): 2.10 mol
- Temperature at point \( A \) (\( T_A \)): 782.0 K
#### Find
- Temperature at point \( D \) (\( T_D \)).
Given that the process from \( A \) to \( D \) is isobaric, we can use the Ideal Gas Law to relate the temperatures and volumes at the two points:
\[ P_2V_A = nRT_A \]
\[ P_2V_D = nRT_D \]
Since the pressures at points \( A \) and \( D \) are the same (\( P_2 \)), we can set up the following proportion:
\[ \frac{V_A}{T_A} = \frac{V_D}{T_D} \]
Given the volumes \( V_A = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b578b6a-d8d1-4da1-b505-9cebf317d6ed%2Fb89095cd-8711-42d0-9b5d-f76d57d43401%2Fb17n6yc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Ideal Gas Cycle in Refrigerators: Temperature Calculation
In this example, we will explore the thermodynamic cycle of an ideal monatomic gas within a refrigerator. The gas amount given is 2.10 moles, and we are asked to determine the temperature at point \( D \) on the presented cycle diagram. It is given that the temperature at point \( A \) is 782.0 K.
#### Diagram Analysis
The diagram displays a Pressure-Volume (P-V) cycle of the gas. The cycle consists of four points labeled A, B, C, and D, and it can be broken down into a series of processes:
- From \( A \) to \( D \): Isobaric (constant pressure) expansion.
- From \( D \) to \( C \): Isochoric (constant volume) cooling.
- From \( C \) to \( B \): Isobaric (constant pressure) compression.
- From \( B \) to \( A \): Isochoric (constant volume) heating.
The key values depicted in the diagram include:
- Pressure \( P_2 \) at points \( A \) and \( D \).
- Pressure \( 1.30 \text{ kPa} \) at points \( B \) and \( C \).
- Volume \( 1.50 \text{ m}^3 \) at points \( A \) and \( B \).
- Volume \( 2.25 \text{ m}^3 \) at points \( C \) and \( D \).
#### Given
- Moles of gas (\( n \)): 2.10 mol
- Temperature at point \( A \) (\( T_A \)): 782.0 K
#### Find
- Temperature at point \( D \) (\( T_D \)).
Given that the process from \( A \) to \( D \) is isobaric, we can use the Ideal Gas Law to relate the temperatures and volumes at the two points:
\[ P_2V_A = nRT_A \]
\[ P_2V_D = nRT_D \]
Since the pressures at points \( A \) and \( D \) are the same (\( P_2 \)), we can set up the following proportion:
\[ \frac{V_A}{T_A} = \frac{V_D}{T_D} \]
Given the volumes \( V_A = 1
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