Question

Calculate the molar entropy of a constant-volume sample of argon at 250K given that it is 154.84 J/K/mol at 298K.

Answer 1

[AI-generated answer — Bot GPT 3.5]

To calculate the molar entropy of a constant-volume sample of argon at 250K using the given value at 298K, we can use the equation:

ΔS = Cp * ln(T2/T1) - R * ln(V2/V1)

Where:
ΔS = Change in molar entropy
Cp = molar heat capacity at constant pressure
R = Gas constant (8.314 J/mol*K)
T1 = initial temperature (298K)
T2 = final temperature (250K)
V1 = initial volume
V2 = final volume

Given that the sample is at constant volume, V2/V1 = 1, so this term drops out of the equation.

Given:
Initial temperature, T1 = 298K
Final temperature, T2 = 250K
Molar entropy at 298K, S1 = 154.84 J/mol*K

Using the equation for molar entropy change:

ΔS = Cp * ln(T2/T1) - R * ln(V2/V1)
ΔS = Cp * ln(250/298) - R * ln(1)
ΔS = Cp * ln(0.8389)

Since we have the molar entropy at 298K (S1), we can express Cp as:

S1 = Cp * ln(T1) - R * ln(V1)

Rearranging the equation to solve for Cp:

Cp = (S1 + R * ln(V1)) / ln(T1)

Substitute values:

Cp = (154.84 + 8.314 * ln(1)) / ln(298) ≈ 31.38 J/mol*K

Now we can calculate the molar entropy at 250K:

ΔS = 31.38 * ln(0.8389) ≈ 3.99 J/mol*K

Therefore, the molar entropy of a constant-volume sample of argon at 250K is approximately 3.99 J/K/mol.