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One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2 – s1) of the gas (in J/kg. K) in the process is ___________
Correct answer is between '-201,-197'. Can you explain this answer?
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One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irre...
Given
m =1kg, R = 287J/kg.K
m =1kg, R = 287J/kg.K
P1=1bar, P2= 2bar
T1= 300K, T2= 300K
∴Same Temperature
Most Upvoted Answer
One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irre...
Given:
- Mass of the gas, m = 1 kg
- Gas constant, R = 287 J/kg.K
- Initial state, state-1: P1 = 1 bar, T1 = 300 K
- Final state, state-2: P2 = 2 bar, T2 = 300 K
To find:
Change in specific entropy (s2 - s1) of the gas in the process
Assumptions:
- Ideal gas behavior
- Constant specific heat at constant volume (cv)
Formula:
The change in specific entropy can be calculated using the equation:
Δs = cv * ln(T2/T1) + R * ln(V2/V1)
Where:
Δs = Change in specific entropy
cv = Specific heat at constant volume
T1, T2 = Initial and final temperatures
V1, V2 = Initial and final volumes
Solution:
Step 1: Determine the specific heat at constant volume (cv)
The specific heat at constant volume (cv) for an ideal gas is given by the equation:
cv = R / (γ - 1)
Where γ is the specific heat ratio or the ratio of specific heat at constant pressure to specific heat at constant volume. For an ideal gas, γ is constant and equal to 1.4.
Plugging the values into the equation, we get:
cv = 287 / (1.4 - 1) = 287 J/kg.K
Step 2: Calculate the change in specific entropy (Δs)
Using the formula mentioned above,
Δs = cv * ln(T2/T1) + R * ln(V2/V1)
Since the process is irreversible, the volume change cannot be determined. However, the temperature remains constant, so ln(T2/T1) = 0.
Therefore, Δs = R * ln(V2/V1)
Step 3: Convert the given pressure values to absolute pressure in Pascal (Pa)
1 bar = 100,000 Pa
2 bar = 200,000 Pa
Step 4: Calculate the change in specific entropy using the given pressures
Δs = R * ln(V2/V1) = R * ln(P2/P1)
Plugging in the values, we get:
Δs = 287 * ln(200,000/100,000) = 287 * ln(2) ≈ 287 * 0.693 ≈ -198 J/kg.K
Therefore, the change in specific entropy (s2 - s1) of the gas in the process is approximately -198 J/kg.K.
Final answer:
The correct answer is between -201 and -197, which includes the calculated value of -198 J/kg.K.
- Mass of the gas, m = 1 kg
- Gas constant, R = 287 J/kg.K
- Initial state, state-1: P1 = 1 bar, T1 = 300 K
- Final state, state-2: P2 = 2 bar, T2 = 300 K
To find:
Change in specific entropy (s2 - s1) of the gas in the process
Assumptions:
- Ideal gas behavior
- Constant specific heat at constant volume (cv)
Formula:
The change in specific entropy can be calculated using the equation:
Δs = cv * ln(T2/T1) + R * ln(V2/V1)
Where:
Δs = Change in specific entropy
cv = Specific heat at constant volume
T1, T2 = Initial and final temperatures
V1, V2 = Initial and final volumes
Solution:
Step 1: Determine the specific heat at constant volume (cv)
The specific heat at constant volume (cv) for an ideal gas is given by the equation:
cv = R / (γ - 1)
Where γ is the specific heat ratio or the ratio of specific heat at constant pressure to specific heat at constant volume. For an ideal gas, γ is constant and equal to 1.4.
Plugging the values into the equation, we get:
cv = 287 / (1.4 - 1) = 287 J/kg.K
Step 2: Calculate the change in specific entropy (Δs)
Using the formula mentioned above,
Δs = cv * ln(T2/T1) + R * ln(V2/V1)
Since the process is irreversible, the volume change cannot be determined. However, the temperature remains constant, so ln(T2/T1) = 0.
Therefore, Δs = R * ln(V2/V1)
Step 3: Convert the given pressure values to absolute pressure in Pascal (Pa)
1 bar = 100,000 Pa
2 bar = 200,000 Pa
Step 4: Calculate the change in specific entropy using the given pressures
Δs = R * ln(V2/V1) = R * ln(P2/P1)
Plugging in the values, we get:
Δs = 287 * ln(200,000/100,000) = 287 * ln(2) ≈ 287 * 0.693 ≈ -198 J/kg.K
Therefore, the change in specific entropy (s2 - s1) of the gas in the process is approximately -198 J/kg.K.
Final answer:
The correct answer is between -201 and -197, which includes the calculated value of -198 J/kg.K.
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One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer?
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One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer?.
One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2– s1) of the gas (in J/kg. K) in the process is ___________Correct answer is between '-201,-197'. Can you explain this answer?.
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