The change in entropy of 2 moles of an ideal gas upon isothermal expansion at 243.6 K from 20 L to the state where pressure becomes 1 atm is
(Given : In 2 = 0.693)

Options

A. 1.385 cal/K
B. `-1.2` cal/K
C. 2.77 cal/K
D. 1.2 cal/K

Answer

To find the change in entropy (ΔS) of 2 moles of an ideal gas during isothermal expansion, we can follow these steps:

### Step 1: Identify the given values
- Number of moles (n) = 2 moles
- Initial volume (V1) = 20 L
- Final pressure (P2) = 1 atm
- Temperature (T) = 243.6 K
- R (gas constant in calorie) = 2 cal/(mol·K)

### Step 2: Use the ideal gas equation to find initial pressure (P1)
The ideal gas equation is given by:
\[ PV = nRT \]

We can rearrange this to find pressure (P):
\[ P = \frac{nRT}{V} \]

Substituting the known values:
\[ P1 = \frac{(2 \, \text{mol}) \times (2 \, \text{cal/(mol·K)}) \times (243.6 \, \text{K})}{20 \, \text{L}} \]

Calculating P1:
\[ P1 = \frac{(2 \times 2 \times 243.6)}{20} \]
\[ P1 = \frac{974.4}{20} \]
\[ P1 = 48.72 \, \text{atm} \]

### Step 3: Use the formula for change in entropy (ΔS)
The change in entropy for an ideal gas during isothermal expansion is given by:
\[ \Delta S = -nR \ln\left(\frac{P2}{P1}\right) \]

### Step 4: Substitute the values into the entropy formula
Substituting the values we have:
\[ \Delta S = -2 \times 2 \times \ln\left(\frac{1 \, \text{atm}}{48.72 \, \text{atm}}\right) \]
\[ \Delta S = -4 \ln\left(\frac{1}{48.72}\right) \]
\[ \Delta S = 4 \ln(48.72) \]

### Step 5: Simplify using the properties of logarithms
Using the property of logarithms:
\[ \ln\left(\frac{1}{x}\right) = -\ln(x) \]
Thus:
\[ \Delta S = 4 \ln(48.72) \]

### Step 6: Calculate the value of ΔS
We can use the approximation:
\[ \ln(48.72) \approx \ln(2) + \ln(24.36) \]
Using the given value of \(\ln(2) = 0.693\):
\[ \Delta S \approx 4 \times 0.693 \]
\[ \Delta S \approx 2.77 \, \text{cal/K} \]

### Final Answer:
The change in entropy (ΔS) is approximately **2.77 cal/K**.

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