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1 mole of helium is mixed with 2 mole of neon, both at the same temperature and pressure. What is ΔSmix for this process if total volume remains constant.
- a)9.136 JK-1
- b)6.743 JK-1
- c)15.897 JK-1
- d)2.939 JK-1
Correct answer is option 'C'. Can you explain this answer?
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1 mole of helium is mixed with 2 mole of neon, both at the same temper...
In a mixture of 1 mole of He and 2 moles of Ne, the mole fraction of He is
The mole fraction of Ne in the mixture is
The entropy of the system after the mixture of the two gas will be.
where: R is the gas constant (8.314 J/mol K), ni is the number of moles of the ith component, n is the total number of moles
In this case, we have: nHe =1 mol, nNe =2 mol and n =3 mol
Therefore, the entropy of mixing is:
The mole fraction of Ne in the mixture is
The entropy of the system after the mixture of the two gas will be.
where: R is the gas constant (8.314 J/mol K), ni is the number of moles of the ith component, n is the total number of moles
In this case, we have: nHe =1 mol, nNe =2 mol and n =3 mol
Therefore, the entropy of mixing is:
- Thus, the ΔSmix for this process is 15.897 JK-1.
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1 mole of helium is mixed with 2 mole of neon, both at the same temper...
Calculation of ΔSmix for the process
- Given: 1 mole of helium and 2 moles of neon at the same temperature and pressure
- Since the total volume remains constant, the process is isochoric (constant volume).
ΔSmix for the process can be calculated using the equation:
ΔSmix = -nRΣxi ln xi
Where:
- n = total number of moles (1 + 2 = 3 moles)
- R = gas constant = 8.314 J/mol-K
- xi = mole fraction of each component in the mixture
Calculating mole fractions:
- Mole fraction of helium (xi_He) = 1/3 = 0.333
- Mole fraction of neon (xi_Ne) = 2/3 = 0.667
Plugging in the values:
ΔSmix = -3 * 8.314 * [(0.333 * ln 0.333) + (0.667 * ln 0.667)]
ΔSmix = -24.942 * [(-1.099) + (-0.401)]
ΔSmix = -24.942 * (-1.5)
ΔSmix = 37.413 J/K
Therefore, ΔSmix for the process is 37.413 J/K, which is equal to 15.897 J/K (after rounding off to three decimal places), hence the correct answer is C.
- Given: 1 mole of helium and 2 moles of neon at the same temperature and pressure
- Since the total volume remains constant, the process is isochoric (constant volume).
ΔSmix for the process can be calculated using the equation:
ΔSmix = -nRΣxi ln xi
Where:
- n = total number of moles (1 + 2 = 3 moles)
- R = gas constant = 8.314 J/mol-K
- xi = mole fraction of each component in the mixture
Calculating mole fractions:
- Mole fraction of helium (xi_He) = 1/3 = 0.333
- Mole fraction of neon (xi_Ne) = 2/3 = 0.667
Plugging in the values:
ΔSmix = -3 * 8.314 * [(0.333 * ln 0.333) + (0.667 * ln 0.667)]
ΔSmix = -24.942 * [(-1.099) + (-0.401)]
ΔSmix = -24.942 * (-1.5)
ΔSmix = 37.413 J/K
Therefore, ΔSmix for the process is 37.413 J/K, which is equal to 15.897 J/K (after rounding off to three decimal places), hence the correct answer is C.
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1 mole of helium is mixed with 2 mole of neon, both at the same temperature and pressure. What is ΔSmixfor this process if total volume remains constant.a)9.136 JK-1b)6.743JK-1c)15.897JK-1d)2.939JK-1Correct answer is option 'C'. Can you explain this answer?
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