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

Answer 1

To calculate the molar entropy of the argon sample at 250K, we can make use of the following equation:

\[ΔS = \int_{T_1}^{T_2} \frac{C_p}{T} dT\]

Given that argon is a monoatomic gas and its heat capacity at constant volume is 3/2R, where R is the gas constant (8.31 J/mol-K), we have:

\[C_v = \frac{3}{2}R = \frac{3}{2} \times 8.31 = 12.46 J/mol-K\]

Therefore, the change in entropy from 298K to 250K is:

\[ΔS = \int_{250}^{298} \frac{12.46}{T} dT\]

\[ΔS = 12.46 \ln\left(\frac{298}{250}\right)\]

\[ΔS = 12.46 \ln\left(1.192\right)\]

\[ΔS = 12.46 \times 0.175\]

\[ΔS = 2.18 J/mol-K\]

To find the molar entropy at 250K:
\[S_{250K} = S_{298K} + ΔS\]

\[S_{250K} = 154.84 + 2.18\]

\[S_{250K} = 157.02 J/mol-K\]

Therefore, the molar entropy of the argon sample at 250K is 157.02 J/mol-K.