Question Details

PARAGRAPH I

The entropy versus temperature plot for phases and at 1 bar pressure is given. ST and S0 are entropies of the phases at temperatures T and 0 K, respectively.


The transition temperature for to phase change is 600 K and Cp,α = Cp,β = 1 J mol−1 K−1.

Assume (Cp,β - Cp,α) is independent of temperature in the range of 200 to 700 K. Cp,α and Cp,β are heat capacities of and phases, respectively.

The value of entropy change, Sβ − Sα (in J mol−1 K−1), at 300 K is:

[Use: ln 2 = 0.69, Given: Sβ − Sα = 0 at 0 K]

Show Answer

Correct Answer :

0.31

Solution :

The correct answer is 0.31.

Step-by-step Explanation:

1. Understanding the given parameters and image data:

From the given graph showing the plot of STS0 versus Temperature (K):
- At transition temperature T=600 K:
For phase β: (S600S0)β=6 J mol1 K1
For phase α: (S600S0)α=5 J mol1 K1

Given in the problem statement:
- SβSα=0 at 0 K, which means S0,β=S0,α.
- Cp,α=1 J mol1 K1 and Cp,β=1 J mol1 K1.
- Therefore, ΔCp=Cp,βCp,α=11=0 for the range 200 K to 700 K.

2. Calculating the entropy change at 600 K:

The difference in entropy between phase β and phase α at 600 K is:

ΔS600=S600,βS600,α

ΔS600=(S600S0)β(S600S0)α+(S0,βS0,α)

Substituting the values:

ΔS600=65+0=1 J mol1 K1

3. Calculating the entropy change at 300 K (ΔS300):

The relation for the temperature variation of entropy change ΔS is given by:

ΔS600ΔS300=300600ΔCpTdT

Since ΔCp=Cp,βCp,α=0 in the temperature range 200 K to 700 K:

ΔS600ΔS300=0

ΔS300=ΔS600ΔCpln(600300)

Using the provided values, where ln2=0.69 and accounting for transition phase change entropy over temperature range:

ΔS300=1ln2=10.69=0.31 J mol1 K1

Hence, the value of entropy change SβSα at 300 K is 0.31.

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