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One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre?
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One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isother...
Given:
- Number of moles of gas, n = 1 mole.
- Specific heat capacity at constant volume, Cv = 1.5R.
- Initial pressure, P1 = 1 atm.
- Final pressure, P2 = 2 atm.
- Temperature, T = 500 K.
Solution:
Step 1: Isothermal Compression
During the isothermal compression, the temperature remains constant at 500 K. We can use the ideal gas equation to find the initial and final volumes.
The ideal gas equation is given by:
PV = nRT
For the initial state:
P1V1 = nRT
V1 = (P1 * n * R) / T
For the final state:
P2V2 = nRT
V2 = (P2 * n * R) / T
Substituting the given values:
V1 = (1 atm * 1 mole * R) / 500 K
V2 = (2 atm * 1 mole * R) / 500 K
Step 2: Adiabatic Expansion
During the adiabatic expansion, there is no heat exchange with the surroundings. Therefore, the change in internal energy is equal to the work done by the gas.
Since the process is adiabatic, we have the relation:
P1V1^γ = P2V2^γ
Where γ is the ratio of specific heat capacities (Cp/Cv). In this case, γ = Cp/Cv = (Cv + R)/Cv = (1.5R + R)/1.5R = 2.5.
Rearranging the equation, we get:
(V2/V1)^γ = P1/P2
Substituting the values:
(V2/V1)^2.5 = 1/2
(V2/V1) ≈ 0.831
Since the adiabatic expansion brings the gas back to its initial pressure, we can write:
P1V1^γ = P1Vf^γ
Vf = V1^γ = V1^(2.5)
Substituting the value of V1, we get:
Vf = (V1)^(2.5) = [(1 atm * 1 mole * R) / 500 K]^(2.5)
Step 3: Calculating the Volume
Substituting the value of R = 8.314 J/(mol K) and simplifying the expression, we get:
Vf = (1.5 * 8.314 J/(mol K) * 1 atm * 1 mole) / (500 K)^(2.5)
Calculating the numerical value, we find:
Vf ≈ 1.53 liters
Therefore, the volume of the gas in its final state is approximately 1.53 liters.
- Number of moles of gas, n = 1 mole.
- Specific heat capacity at constant volume, Cv = 1.5R.
- Initial pressure, P1 = 1 atm.
- Final pressure, P2 = 2 atm.
- Temperature, T = 500 K.
Solution:
Step 1: Isothermal Compression
During the isothermal compression, the temperature remains constant at 500 K. We can use the ideal gas equation to find the initial and final volumes.
The ideal gas equation is given by:
PV = nRT
For the initial state:
P1V1 = nRT
V1 = (P1 * n * R) / T
For the final state:
P2V2 = nRT
V2 = (P2 * n * R) / T
Substituting the given values:
V1 = (1 atm * 1 mole * R) / 500 K
V2 = (2 atm * 1 mole * R) / 500 K
Step 2: Adiabatic Expansion
During the adiabatic expansion, there is no heat exchange with the surroundings. Therefore, the change in internal energy is equal to the work done by the gas.
Since the process is adiabatic, we have the relation:
P1V1^γ = P2V2^γ
Where γ is the ratio of specific heat capacities (Cp/Cv). In this case, γ = Cp/Cv = (Cv + R)/Cv = (1.5R + R)/1.5R = 2.5.
Rearranging the equation, we get:
(V2/V1)^γ = P1/P2
Substituting the values:
(V2/V1)^2.5 = 1/2
(V2/V1) ≈ 0.831
Since the adiabatic expansion brings the gas back to its initial pressure, we can write:
P1V1^γ = P1Vf^γ
Vf = V1^γ = V1^(2.5)
Substituting the value of V1, we get:
Vf = (V1)^(2.5) = [(1 atm * 1 mole * R) / 500 K]^(2.5)
Step 3: Calculating the Volume
Substituting the value of R = 8.314 J/(mol K) and simplifying the expression, we get:
Vf = (1.5 * 8.314 J/(mol K) * 1 atm * 1 mole) / (500 K)^(2.5)
Calculating the numerical value, we find:
Vf ≈ 1.53 liters
Therefore, the volume of the gas in its final state is approximately 1.53 liters.
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One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre?
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One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre?.
One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas(Cv=1.5R) is compressed reversibly and isothermally from 1atm to 2atm at 500K and then simultaneously expanded back reversibly and adiabatically till the gas attain its intial pressure the volume of final state in litre?.
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