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2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by
- a)–Rln2
- b)0
- c)Rln2
- d)Rln4
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
2 moles of oxygen are mixed adiabatically with another 2 moles of oxyg...
Remember if we mix 2 mole of oxygen with another 2 mole of other gas the volume will be doubled for first and second constituents 
Total Entropy change = 4Rln2 So, Entropy change per mole=Rln2. And it is due to diffusion of one gas into another.
Most Upvoted Answer
2 moles of oxygen are mixed adiabatically with another 2 moles of oxyg...
Option B is correct. For the case of mixing of same gases at same P and T the entropy change will be zero. We can't distinguish between the different molecules of oxygen here to find the mole fraction and put it in the standard equation. So the entropy change is zero in such cases.
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2 moles of oxygen are mixed adiabatically with another 2 moles of oxyg...
Given:
- 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing chamber.
- The final total pressure and temperature of the mixture become the same as those of the individual constituents at their initial states.
- The universal gas constant is given as R.
To Find:
The change in entropy due to mixing, per mole of oxygen.
Solution:
Step 1: Understanding Adiabatic Mixing
When two gases are mixed adiabatically, there is no exchange of heat with the surroundings. Therefore, the process can be considered reversible and adiabatic. In an adiabatic process, the change in entropy can be determined using the equation:
ΔS = Cv * ln(T2/T1) + R * ln(V2/V1)
Where:
ΔS = Change in entropy
Cv = Specific heat capacity at constant volume
T1, T2 = Initial and final temperatures
V1, V2 = Initial and final volumes
R = Universal gas constant
Step 2: Analyzing the Problem
In this scenario, 2 moles of oxygen are mixed with another 2 moles of oxygen. Since the total pressure and temperature of the mixture become the same as the individual constituents at their initial states, we can assume that the initial pressures and temperatures are the same.
Step 3: Applying the Adiabatic Mixing Equation
Since the initial pressures and temperatures are the same, we can assume that T1 = T2 and P1 = P2. Therefore, ln(T2/T1) and ln(V2/V1) become ln(1) = 0.
ΔS = Cv * ln(T2/T1) + R * ln(V2/V1)
ΔS = Cv * 0 + R * 0
ΔS = 0
Step 4: Final Answer
The change in entropy due to mixing, per mole of oxygen, is 0 (option B).
Conclusion:
When 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen, the change in entropy per mole of oxygen is 0. This means that there is no change in entropy during the mixing process.
- 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing chamber.
- The final total pressure and temperature of the mixture become the same as those of the individual constituents at their initial states.
- The universal gas constant is given as R.
To Find:
The change in entropy due to mixing, per mole of oxygen.
Solution:
Step 1: Understanding Adiabatic Mixing
When two gases are mixed adiabatically, there is no exchange of heat with the surroundings. Therefore, the process can be considered reversible and adiabatic. In an adiabatic process, the change in entropy can be determined using the equation:
ΔS = Cv * ln(T2/T1) + R * ln(V2/V1)
Where:
ΔS = Change in entropy
Cv = Specific heat capacity at constant volume
T1, T2 = Initial and final temperatures
V1, V2 = Initial and final volumes
R = Universal gas constant
Step 2: Analyzing the Problem
In this scenario, 2 moles of oxygen are mixed with another 2 moles of oxygen. Since the total pressure and temperature of the mixture become the same as the individual constituents at their initial states, we can assume that the initial pressures and temperatures are the same.
Step 3: Applying the Adiabatic Mixing Equation
Since the initial pressures and temperatures are the same, we can assume that T1 = T2 and P1 = P2. Therefore, ln(T2/T1) and ln(V2/V1) become ln(1) = 0.
ΔS = Cv * ln(T2/T1) + R * ln(V2/V1)
ΔS = Cv * 0 + R * 0
ΔS = 0
Step 4: Final Answer
The change in entropy due to mixing, per mole of oxygen, is 0 (option B).
Conclusion:
When 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen, the change in entropy per mole of oxygen is 0. This means that there is no change in entropy during the mixing process.
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2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by a)–Rln2b)0c)Rln2d)Rln4Correct answer is option 'B'. Can you explain this answer?
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