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Two moles of a monoatomic perfect gas initially 4.0 bars and 470C undergoes reversible expansion in an insulated container. The temperature at which the pressure reduces to 3.0 bar is:
- a)200 K
- b)285 K
- c)310 K
- d)320 K
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Two moles of a monoatomic perfect gas initially 4.0 bars and 470C unde...
Given:
- Two moles of a monoatomic perfect gas
- Initial pressure (P1) = 4.0 bar
- Initial temperature (T1) = 470C
- Final pressure (P2) = 3.0 bar
- Process is reversible and adiabatic (insulated container)
To find: Final temperature (T2)
Formula used:
- For reversible adiabatic process, PV^γ = constant, where γ = Cp/Cv for a monoatomic gas is 5/3
- Also, PV = nRT (Ideal gas equation)
Solution:
1. Using the ideal gas equation, we can find the initial volume (V1) of the gas:
PV = nRT
V1 = nRT1/P1 = 2 x 8.314 x (470+273)/4.0 = 0.078 m3
2. Using the initial pressure and volume, we can find the initial value of PV^γ:
PV^γ = (4.0 bar x 0.078 m3)^5/3 = 1.10 x 10^6 bar3/K5/3
3. Since the process is reversible adiabatic, PV^γ remains constant:
P1V1^γ = P2V2^γ
4. Using the final pressure (P2) and the constant value of PV^γ, we can find the final volume (V2):
V2 = (P1V1^γ/P2)^(1/γ) = (1.10 x 10^6 bar3/K5/3 / 3.0 bar)^(1/5/3) = 0.105 m3
5. Using the final volume and the ideal gas equation, we can find the final temperature (T2):
PV = nRT
T2 = P2V2/nR = 3.0 bar x 0.105 m3 / 2 x 8.314 = 285 K
Therefore, the final temperature at which the pressure reduces to 3.0 bar is 285 K, which is option B.
- Two moles of a monoatomic perfect gas
- Initial pressure (P1) = 4.0 bar
- Initial temperature (T1) = 470C
- Final pressure (P2) = 3.0 bar
- Process is reversible and adiabatic (insulated container)
To find: Final temperature (T2)
Formula used:
- For reversible adiabatic process, PV^γ = constant, where γ = Cp/Cv for a monoatomic gas is 5/3
- Also, PV = nRT (Ideal gas equation)
Solution:
1. Using the ideal gas equation, we can find the initial volume (V1) of the gas:
PV = nRT
V1 = nRT1/P1 = 2 x 8.314 x (470+273)/4.0 = 0.078 m3
2. Using the initial pressure and volume, we can find the initial value of PV^γ:
PV^γ = (4.0 bar x 0.078 m3)^5/3 = 1.10 x 10^6 bar3/K5/3
3. Since the process is reversible adiabatic, PV^γ remains constant:
P1V1^γ = P2V2^γ
4. Using the final pressure (P2) and the constant value of PV^γ, we can find the final volume (V2):
V2 = (P1V1^γ/P2)^(1/γ) = (1.10 x 10^6 bar3/K5/3 / 3.0 bar)^(1/5/3) = 0.105 m3
5. Using the final volume and the ideal gas equation, we can find the final temperature (T2):
PV = nRT
T2 = P2V2/nR = 3.0 bar x 0.105 m3 / 2 x 8.314 = 285 K
Therefore, the final temperature at which the pressure reduces to 3.0 bar is 285 K, which is option B.
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Community Answer
Two moles of a monoatomic perfect gas initially 4.0 bars and 470C unde...
Given information:
- Two moles of a monoatomic perfect gas
- Initial pressure (P1) = 4.0 bar
- Initial temperature (T1) = 470°C
- Final pressure (P2) = 3.0 bar
- The process is reversible and adiabatic
To find: The final temperature (T2) at which the pressure reduces to 3.0 bar.
Formula used:
For an adiabatic process, PVγ is constant, where P is the pressure, V is the volume, and γ is the ratio of specific heats (Cp/Cv) for the gas. For a monoatomic perfect gas, γ = 5/3.
Calculation:
Using the formula PVγ = constant, we can write:
P1V1γ = P2V2γ
Since the gas is monoatomic, we can use the ideal gas law to relate the pressure, volume, and temperature:
PV = nRT
where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
Substituting this expression for V in the first equation and simplifying, we get:
(P1/P2) = (T2/T1)^(γ-1)
(4.0/3.0) = (T2/743)^(2/3)
(T2/743) = (3/4)^(3/2)
T2 = 285 K
Therefore, the final temperature at which the pressure reduces to 3.0 bar is 285 K.
Answer: b) 285 K
- Two moles of a monoatomic perfect gas
- Initial pressure (P1) = 4.0 bar
- Initial temperature (T1) = 470°C
- Final pressure (P2) = 3.0 bar
- The process is reversible and adiabatic
To find: The final temperature (T2) at which the pressure reduces to 3.0 bar.
Formula used:
For an adiabatic process, PVγ is constant, where P is the pressure, V is the volume, and γ is the ratio of specific heats (Cp/Cv) for the gas. For a monoatomic perfect gas, γ = 5/3.
Calculation:
Using the formula PVγ = constant, we can write:
P1V1γ = P2V2γ
Since the gas is monoatomic, we can use the ideal gas law to relate the pressure, volume, and temperature:
PV = nRT
where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
Substituting this expression for V in the first equation and simplifying, we get:
(P1/P2) = (T2/T1)^(γ-1)
(4.0/3.0) = (T2/743)^(2/3)
(T2/743) = (3/4)^(3/2)
T2 = 285 K
Therefore, the final temperature at which the pressure reduces to 3.0 bar is 285 K.
Answer: b) 285 K
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Two moles of a monoatomic perfect gas initially 4.0 bars and 470C undergoes reversible expansion in an insulated container. The temperature at which the pressure reduces to 3.0 bar is:a)200 Kb)285 Kc)310 Kd)320 KCorrect answer is option 'B'. Can you explain this answer?
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Two moles of a monoatomic perfect gas initially 4.0 bars and 470C undergoes reversible expansion in an insulated container. The temperature at which the pressure reduces to 3.0 bar is:a)200 Kb)285 Kc)310 Kd)320 KCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Two moles of a monoatomic perfect gas initially 4.0 bars and 470C undergoes reversible expansion in an insulated container. The temperature at which the pressure reduces to 3.0 bar is:a)200 Kb)285 Kc)310 Kd)320 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of a monoatomic perfect gas initially 4.0 bars and 470C undergoes reversible expansion in an insulated container. The temperature at which the pressure reduces to 3.0 bar is:a)200 Kb)285 Kc)310 Kd)320 KCorrect answer is option 'B'. Can you explain this answer?.
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