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For the process, 1 Ar (300 K, 1 bar) rar...

For the process, 1 Ar (300 K, 1 bar) `rarr` 1 Ar (200 K, 10 bar), assuming ideal gas behaviour, the change in molar entropy is

A

`-27.58J//K//mol`

B

`+27.58J//K//mol`

C

`-24.28J//K//mol`

D

`+24.28J//K//mol`

Text Solution

AI Generated Solution

The correct Answer is:
To find the change in molar entropy (ΔS) for the process where 1 mole of Argon gas is changed from 300 K and 1 bar to 200 K and 10 bar, we can use the formula for the change in entropy when both temperature and pressure change: \[ \Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right) \] ### Step 1: Identify the parameters - \( n = 1 \) mole (since we are dealing with 1 Ar) - \( T_1 = 300 \, K \) - \( T_2 = 200 \, K \) - \( P_1 = 1 \, bar \) - \( P_2 = 10 \, bar \) ### Step 2: Determine the value of \( C_p \) for Argon For a monoatomic ideal gas like Argon, the molar heat capacity at constant pressure \( C_p \) is given by: \[ C_p = \frac{5}{2} R \] Where \( R \) (the gas constant) is approximately \( 8.314 \, J/(mol \cdot K) \). Calculating \( C_p \): \[ C_p = \frac{5}{2} \times 8.314 = 20.835 \, J/(mol \cdot K) \] ### Step 3: Substitute values into the entropy change formula Now we can substitute the values into the entropy change formula: \[ \Delta S = 1 \times 20.835 \ln\left(\frac{200}{300}\right) - 1 \times 8.314 \ln\left(\frac{10}{1}\right) \] ### Step 4: Calculate the logarithmic terms Calculating the logarithmic terms: 1. For the temperature term: \[ \ln\left(\frac{200}{300}\right) = \ln\left(\frac{2}{3}\right) \approx -0.4055 \] 2. For the pressure term: \[ \ln\left(\frac{10}{1}\right) = \ln(10) \approx 2.3026 \] ### Step 5: Substitute the logarithmic values back into the equation Now substituting these values back into the equation: \[ \Delta S = 20.835 \times (-0.4055) - 8.314 \times 2.3026 \] Calculating each term: 1. First term: \[ 20.835 \times (-0.4055) \approx -8.447 \] 2. Second term: \[ 8.314 \times 2.3026 \approx 19.157 \] ### Step 6: Combine the results Now combine the results: \[ \Delta S = -8.447 - 19.157 \approx -27.604 \, J/(mol \cdot K) \] ### Final Result The change in molar entropy is approximately: \[ \Delta S \approx -27.6 \, J/(mol \cdot K) \]
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