Question

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

Answer 1

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 2

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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