Chemistry, asked by gitangli2251, 10 months ago

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

Answered by brainlyrealhead
3

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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Answered by Glitterash
1

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

#SPJ3

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