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The entropy change when an ideal gas und...

The entropy change when an ideal gas under atmospheric condition at room temperature is allowed to expand from 0.5 L to 1.0 L and also is simultaneoulsy heated to 373 L will be
`("Given : "C_("v, m")="12.50 J K"^(-1)"mol"^(-1) and log 1.25=0.1)`

A

`0.18JK^(-1)`

B

`0.36JK^(-1)`

C

`0.90JK^(-1)`

D

`0.72JK^(-1)`

Text Solution

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The correct Answer is:
To solve the problem of calculating the entropy change when an ideal gas expands and is heated, we can follow these steps: ### Step 1: Identify Initial Conditions - Initial volume (V1) = 0.5 L - Final volume (V2) = 1.0 L - Initial temperature (T1) = 298 K (room temperature) - Final temperature (T2) = 373 K - Pressure (P) = 1 atm ### Step 2: Calculate the Number of Moles (n) Using the ideal gas law, \( PV = nRT \): \[ n = \frac{PV}{RT} \] Where: - \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \) Substituting the values: \[ n = \frac{(1 \, \text{atm}) \times (0.5 \, \text{L})}{(0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (298 \, \text{K})} \] Calculating this gives: \[ n = \frac{0.5}{24.4756} \approx 0.0204 \, \text{mol} \] ### Step 3: Use the Entropy Change Formula The entropy change (\( \Delta S \)) for an ideal gas can be calculated using the formula: \[ \Delta S = nC_v \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right) \] Where: - \( C_v = 12.5 \, \text{J K}^{-1} \text{mol}^{-1} \) ### Step 4: Calculate Each Component 1. **Temperature Change Component**: \[ nC_v \ln\left(\frac{T_2}{T_1}\right) = 0.0204 \times 12.5 \times \ln\left(\frac{373}{298}\right) \] First, calculate \( \frac{373}{298} \approx 1.25 \). Then, using \( \ln(1.25) \approx 0.2231 \): \[ nC_v \ln\left(\frac{T_2}{T_1}\right) \approx 0.0204 \times 12.5 \times 0.2231 \approx 0.0571 \, \text{J K}^{-1} \] 2. **Volume Change Component**: \[ nR \ln\left(\frac{V_2}{V_1}\right) = 0.0204 \times 0.082 \times \ln\left(\frac{1.0}{0.5}\right) \] Since \( \frac{1.0}{0.5} = 2 \), we have \( \ln(2) \approx 0.693 \): \[ nR \ln\left(\frac{V_2}{V_1}\right) \approx 0.0204 \times 0.082 \times 0.693 \approx 0.0011 \, \text{J K}^{-1} \] ### Step 5: Combine the Components Now, sum both components to find the total entropy change: \[ \Delta S \approx 0.0571 + 0.0011 \approx 0.0582 \, \text{J K}^{-1} \] ### Step 6: Final Calculation After rounding, we find: \[ \Delta S \approx 0.058 \, \text{J K}^{-1} \] ### Conclusion The entropy change when the ideal gas expands and is heated under the given conditions is approximately \( 0.058 \, \text{J K}^{-1} \). ---
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