Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.2870 kJ/(kg-K). (The internal energy can be determined by the equation AU=m-cy (T2-T1), where cy= 0.7180 kJ/(kg-K) is the specific heat at the constant volume.) Calculate the final state pressure p2. Ideal gas T1 P1 V1 FISE State 1 Evacuated (kPa) Ideal gas T2=? P2=? V2=2V1 State 2
Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.2870 kJ/(kg-K). (The internal energy can be determined by the equation AU=m-cy (T2-T1), where cy= 0.7180 kJ/(kg-K) is the specific heat at the constant volume.) Calculate the final state pressure p2. Ideal gas T1 P1 V1 FISE State 1 Evacuated (kPa) Ideal gas T2=? P2=? V2=2V1 State 2
Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.2870 kJ/(kg·K). (The internal energy can be determined by the equation ΔU=m·cv·(T2-T1), where cv = 0.7180 kJ/(kg·K) is the specific heat at the constant volume.)
Calculate the final state pressure p2.__________ (kPa)
![In the illustrated scenario, an insulated rigid tank is separated into two equal compartments by a partition. Initially, one of these compartments contains an ideal gas, while the other is evacuated (empty). The partition is subsequently removed, allowing the gas to expand into the entire tank.
In the initial state (State 1), the gas properties are defined as follows:
- Temperature \( T1 = 300.00 \, K \)
- Pressure \( P1 = 600.00 \, kPa \)
- Volume \( V1 \) (occupying half of the tank)
- Mass \( m = 4.00 \, kg \)
The gas constant provided is \( R = 0.2870 \, \text{kJ/(kg·K)} \).
The specific heat at constant volume is \( c_v = 0.7180 \, \text{kJ/(kg·K)} \).
The change in internal energy can be calculated using the formula:
\[
\Delta U = m \cdot c_v \cdot (T2 - T1)
\]
In the subsequent state (State 2), after the partition is removed, the variables are:
- Temperature \( T2 \) (to be determined)
- Pressure \( P2 \) (to be calculated)
- Volume \( V2 = 2V1 \) (since the gas fills the entire tank)
The task is to calculate the final state pressure \( P2 \) in \( kPa \).
**Diagram Explanation:**
The diagram shows two states visually:
- **State 1:** The left compartment contains the ideal gas, while the right is evacuated.
- **State 2:** The partition is removed, and the gas expands to fill both compartments equally.
This setup is often used to demonstrate the principles of gas expansion and the application of the first law of thermodynamics in closed systems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35813974-6b68-4217-bd4c-21bf1a217db1%2Fe6efe12d-8b87-4f30-a86c-a141076c1df5%2Fjdn4avh_processed.png&w=3840&q=75)
Transcribed Image Text:In the illustrated scenario, an insulated rigid tank is separated into two equal compartments by a partition. Initially, one of these compartments contains an ideal gas, while the other is evacuated (empty). The partition is subsequently removed, allowing the gas to expand into the entire tank.
In the initial state (State 1), the gas properties are defined as follows:
- Temperature \( T1 = 300.00 \, K \)
- Pressure \( P1 = 600.00 \, kPa \)
- Volume \( V1 \) (occupying half of the tank)
- Mass \( m = 4.00 \, kg \)
The gas constant provided is \( R = 0.2870 \, \text{kJ/(kg·K)} \).
The specific heat at constant volume is \( c_v = 0.7180 \, \text{kJ/(kg·K)} \).
The change in internal energy can be calculated using the formula:
\[
\Delta U = m \cdot c_v \cdot (T2 - T1)
\]
In the subsequent state (State 2), after the partition is removed, the variables are:
- Temperature \( T2 \) (to be determined)
- Pressure \( P2 \) (to be calculated)
- Volume \( V2 = 2V1 \) (since the gas fills the entire tank)
The task is to calculate the final state pressure \( P2 \) in \( kPa \).
**Diagram Explanation:**
The diagram shows two states visually:
- **State 1:** The left compartment contains the ideal gas, while the right is evacuated.
- **State 2:** The partition is removed, and the gas expands to fill both compartments equally.
This setup is often used to demonstrate the principles of gas expansion and the application of the first law of thermodynamics in closed systems.
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