The entropy of mixing of one mole of oxygen gas and two moles of hydrogen gas assuming that no chemical reaction occurs and the gas mixture behaves ideally
Answers
Explanation:
# The entropy of mixing one mole of oxygen gas and two moles of hydrogen gas :
• Figure 1 : ( left ) Two gases A and B in their respective volumes and ( right ) . A homogenous
mixture of gases A and B .
To calculate the entropy change , let us treat this mixing as two separate gas expansions , one for gas A and another for B . From the Statistical definition of entropy , we know that
∆ S = nRIn V2
V1 (1)
• Now for each gas , the volume V1 is the initial volume of the gas , and V2 is the final volume , which is both the gases combined , VA + VB . So for the two separate gas expansions ,
∆ SA = nARIn VA + VB
VA (2)
∆ SB = nBRIn VA + VB
VB (3)
• So to find the total entropy change for both these processes , because they are happening at the same time , we simply add the two changes in entropy together ,
∆ mix S = ASa + ∆ SB =
nARIn VA + VB
VA + nBRIn (4)
• Recalling the ideal gas law , PV = nRT , we see that the volume is directly proportional to the number of moles ( Avogadro's law ) , and since we know the number of moles we can substitute this for the volume :
∆ mix S = nARIn nA + nB
nA + nBRIn nA + nB
nB (5)
• Representing the equation for the entropy change of mixing . This equation is also commonly written with the total number of moles :
∆ mix S = -n R ( XAIn XA + XB In XB ) (6)
• Where the total number of moles is :
n = nA + nB
Given:
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Answer:
34.0 J/K/mol
Explanation:
Given:
number or moles of oxygen = 1
number of moles of hydrogen = 2
To Find:
Entropy change = ?
Solution:
Entropy change of a mixture [ΔS] is given by
Δ₍mix₎S = ΔS₁+ΔS₂
= n₁Rln [n₁+n₂]/n₁ + n₂Rln [n₁+n₂]/n₂
ΔS₁ = entropy change for oxygen
ΔS₂ = entropy change for hydrogen
n₁ = number of moles of oxygen
n₂ = number of moles of hydrogen
X₁ = mole fraction of oxygen
X₂ = mole fraction of hydrogen
R = 8.313
X₁ = [1+2]/1 = 3
X₂ = [1+2]/2 = 3/2
Δ₍mix₎S = -n₁RlnX₁ -n₂RlnX₂
Δ₍mix₎S = -3Rln3-2Rln3/2
Δ₍mix₎S = -R[3ln3+2ln3/2]
Δ₍mix₎S = -R[3.29+0.81]
Δ₍mix₎S = 34.0J/K/mol
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