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Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)
Correct answer is '22.0 to 24.0'. Can you explain this answer?
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Two moles of an ideal gas X and two moles of an ideal gas Y, initially...
Since the process is isothermal and isobaric, we know that the Gibbs free energy change (ΔG) is equal to the enthalpy change (ΔH) minus the temperature times the entropy change (ΔS):
ΔG = ΔH - TΔS
Since the process is ideal, we can use the ideal gas law to find the initial and final states of the gases:
PV = nRT
For gas X:
P1V1 = n1RT
For gas Y:
P2V2 = n2RT
Since the gases are initially at the same temperature and pressure, we can assume that P1 = P2 and T1 = T2. Therefore:
V1 = n1RT1/P1 = n1RT2/P2 = V2
And:
n1 + n2 = 4
After mixing, the final pressure and temperature will also be the same as the initial pressure and temperature. Therefore, the final volume will be:
Vf = V1 + V2 = 2V1
The final number of moles of each gas will also be the same:
nfX = nfY = 2
Using the ideal gas law, we can find the final volume of each gas:
nXRT = P2Vf/2
nYRT = P2Vf/2
Substituting Vf = 2V1 and P2 = P1:
nXRT = nYRT = 2P1V1
Therefore:
nX = nY = 2
The final entropy change can be found using the formula:
ΔS = -ΔG/T
Since the process is isothermal, T is constant and can be factored out:
ΔS = -ΔG/T = -(ΔH - TΔS)/T
Substituting ΔH = 0 (since the process is isobaric and no energy is exchanged):
ΔS = -(-TΔS)/T = ΔS
The entropy change is therefore:
ΔS = R ln (Vf/V1) + 2R ln 2
ΔS = R ln (2V1/V1) + 2R ln 2
ΔS = 2R ln 2
Substituting R = 8.314 J/(mol K):
ΔS = 2(8.314 J/(mol K)) ln 2
ΔS = 11.6 J/K
ΔG = ΔH - TΔS
Since the process is ideal, we can use the ideal gas law to find the initial and final states of the gases:
PV = nRT
For gas X:
P1V1 = n1RT
For gas Y:
P2V2 = n2RT
Since the gases are initially at the same temperature and pressure, we can assume that P1 = P2 and T1 = T2. Therefore:
V1 = n1RT1/P1 = n1RT2/P2 = V2
And:
n1 + n2 = 4
After mixing, the final pressure and temperature will also be the same as the initial pressure and temperature. Therefore, the final volume will be:
Vf = V1 + V2 = 2V1
The final number of moles of each gas will also be the same:
nfX = nfY = 2
Using the ideal gas law, we can find the final volume of each gas:
nXRT = P2Vf/2
nYRT = P2Vf/2
Substituting Vf = 2V1 and P2 = P1:
nXRT = nYRT = 2P1V1
Therefore:
nX = nY = 2
The final entropy change can be found using the formula:
ΔS = -ΔG/T
Since the process is isothermal, T is constant and can be factored out:
ΔS = -ΔG/T = -(ΔH - TΔS)/T
Substituting ΔH = 0 (since the process is isobaric and no energy is exchanged):
ΔS = -(-TΔS)/T = ΔS
The entropy change is therefore:
ΔS = R ln (Vf/V1) + 2R ln 2
ΔS = R ln (2V1/V1) + 2R ln 2
ΔS = 2R ln 2
Substituting R = 8.314 J/(mol K):
ΔS = 2(8.314 J/(mol K)) ln 2
ΔS = 11.6 J/K
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Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer?
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Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer?.
Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of an ideal gas X and two moles of an ideal gas Y, initially at the same temperature and pressure, are mixed under isothermal-isobaric condition. The entropy change on mixing is __________J K–1. (Up to one decimal place. Use R = 8.31 J K–1 mol–1)Correct answer is '22.0 to 24.0'. Can you explain this answer?.
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