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)`

Options

A. `0.18JK^(-1)`
B. `0.36JK^(-1)`
C. `0.90JK^(-1)`
D. `0.72JK^(-1)`

Answer

To solve the problem of calculating the entropy change when an ideal gas expands and is heated, we can follow these steps:

### Step 1: Determine the Initial Conditions
- We know the initial volume \( V_1 = 0.5 \, \text{L} \).
- The final volume \( V_2 = 1.0 \, \text{L} \).
- The initial temperature \( T_1 = 298 \, \text{K} \) (room temperature).
- The final temperature \( T_2 = 373 \, \text{K} \) (heating condition).

### Step 2: Calculate the Number of Moles
Using the ideal gas equation:
\[ PV = nRT \]
Assuming atmospheric pressure \( P = 1 \, \text{atm} = 101.3 \, \text{kPa} \) and using the universal gas constant \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \):
\[ n = \frac{PV}{RT} = \frac{(1 \, \text{atm})(0.5 \, \text{L})}{(0.082 \, \text{L atm K}^{-1} \text{mol}^{-1})(298 \, \text{K})} \]
Calculating this gives:
\[ n = \frac{0.5}{24.4756} \approx 0.0204 \, \text{mol} \]

### Step 3: Calculate the Entropy Change
The total 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:
- \( C_V = 12.50 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)

Substituting the values:
1. Calculate \( nC_V \ln\left(\frac{T_2}{T_1}\right) \):
\[ nC_V \ln\left(\frac{373}{298}\right) = 0.0204 \times 12.50 \times \ln(1.2517) \]
Using \( \ln(1.2517) \approx 0.2231 \):
\[ = 0.0204 \times 12.50 \times 0.2231 \approx 0.0571 \, \text{J K}^{-1} \]

2. Calculate \( nR \ln\left(\frac{V_2}{V_1}\right) \):
\[ nR \ln\left(\frac{1.0}{0.5}\right) = 0.0204 \times 8.314 \times \ln(2) \]
Using \( \ln(2) \approx 0.6931 \):
\[ = 0.0204 \times 8.314 \times 0.6931 \approx 0.0940 \, \text{J K}^{-1} \]

### Step 4: Combine the Results
Adding both contributions to find the total entropy change:
\[ \Delta S = 0.0571 + 0.0940 \approx 0.1511 \, \text{J K}^{-1} \]

### Final Answer
The total entropy change when the gas expands and is heated is approximately \( 0.1511 \, \text{J K}^{-1} \).

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