2.0 moles of Helium gas that initially occupies a volume of 5.0 L at a pressure of 100 kPa undergoes the following thermodynamic heat-engine cycle described below: What is the efficiency of the engine? P(kPa) 4 O a. 26% b.14% C. 10% d. 22% e. 18% 200- 100+ Q12 1 5.0 Q23 Work Q41 3 Q34 20 V(L)
2.0 moles of Helium gas that initially occupies a volume of 5.0 L at a pressure of 100 kPa undergoes the following thermodynamic heat-engine cycle described below: What is the efficiency of the engine? P(kPa) 4 O a. 26% b.14% C. 10% d. 22% e. 18% 200- 100+ Q12 1 5.0 Q23 Work Q41 3 Q34 20 V(L)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
-
2.0 moles of Helium gas that initially occupies a volume of 5.0 L at a pressure of 100 kPa undergoes the following
thermodynamic heat-engine cycle described below: What is the efficiency of the engine?
a. 26%
b. 14%
c. 10%
d. 22%
e. 18%
![**Thermodynamic Heat-Engine Cycle Analysis**
In this example, 2.0 moles of helium gas initially occupy a volume of 5.0 L at a pressure of 100 kPa. The gas undergoes a thermodynamic heat-engine cycle depicted in the pressure-volume (P-V) diagram below. The question posed is: What is the efficiency of the engine?
**P-V Diagram Explanation:**
- The diagram is a rectangle with points labeled 1, 2, 3, and 4 representing different states in the cycle.
- The x-axis represents volume (V) in liters (L), ranging from 0 to 20 L, with significant points at 5.0 L and 20 L.
- The y-axis represents pressure (P) in kilopascals (kPa), ranging from 0 to 300 kPa, with key pressure levels at 100 kPa and 200 kPa.
- The cycle moves clockwise through the points:
- 1 to 2 (horizontal, constant volume)
- 2 to 3 (vertical, constant pressure)
- 3 to 4 (horizontal, constant volume)
- 4 to 1 (vertical, constant pressure)
- Heat transfers are labeled for each segment: \(Q_{12}\), \(Q_{23}\), \(Q_{34}\), and \(Q_{41}\).
**Efficiency Options (a-e):**
Select the efficiency of the engine from the following options:
- a. 26%
- b. 14%
- c. 10%
- d. 22%
- e. 18%
To solve for efficiency, students would typically use the thermodynamic principles to calculate work done and heat exchange, applying the efficiency formula:
\[ \text{Efficiency} = \frac{\text{Work output}}{\text{Heat input}} \times 100\% \]
This diagram is an illustrative tool for understanding the mechanical work and heat transfer in a thermodynamic cycle involving an ideal gas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35813974-6b68-4217-bd4c-21bf1a217db1%2F8cc56aa0-6523-4adc-8d6e-15df1d2f56bb%2Frtqnowe_processed.png&w=3840&q=75)
Transcribed Image Text:**Thermodynamic Heat-Engine Cycle Analysis**
In this example, 2.0 moles of helium gas initially occupy a volume of 5.0 L at a pressure of 100 kPa. The gas undergoes a thermodynamic heat-engine cycle depicted in the pressure-volume (P-V) diagram below. The question posed is: What is the efficiency of the engine?
**P-V Diagram Explanation:**
- The diagram is a rectangle with points labeled 1, 2, 3, and 4 representing different states in the cycle.
- The x-axis represents volume (V) in liters (L), ranging from 0 to 20 L, with significant points at 5.0 L and 20 L.
- The y-axis represents pressure (P) in kilopascals (kPa), ranging from 0 to 300 kPa, with key pressure levels at 100 kPa and 200 kPa.
- The cycle moves clockwise through the points:
- 1 to 2 (horizontal, constant volume)
- 2 to 3 (vertical, constant pressure)
- 3 to 4 (horizontal, constant volume)
- 4 to 1 (vertical, constant pressure)
- Heat transfers are labeled for each segment: \(Q_{12}\), \(Q_{23}\), \(Q_{34}\), and \(Q_{41}\).
**Efficiency Options (a-e):**
Select the efficiency of the engine from the following options:
- a. 26%
- b. 14%
- c. 10%
- d. 22%
- e. 18%
To solve for efficiency, students would typically use the thermodynamic principles to calculate work done and heat exchange, applying the efficiency formula:
\[ \text{Efficiency} = \frac{\text{Work output}}{\text{Heat input}} \times 100\% \]
This diagram is an illustrative tool for understanding the mechanical work and heat transfer in a thermodynamic cycle involving an ideal gas.
Expert Solution
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Step 1
Given data :-
For helium
Moles = 2
Initial volume = 5L
Initial pressure = 100kPa
Brief introduction:-
find the temperature at state 1,2 and 3.
Then find total heat supplied .find efficiency by dividing net work by total heat supplied.
Step by step
Solved in 4 steps with 3 images
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