In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change `("Cal K"^(-1))` in the process is -

Options

A. `+1.39`
B. `-1.39`
C. `+1.59`
D. `-1.59`

Answer

To solve the problem, we need to calculate the total entropy change during the isothermal expansion of an ideal gas. Here are the steps to arrive at the solution:

### Step 1: Identify the Given Data
- Temperature (T) = 300 K
- Number of moles (n) = 1 mole
- Initial pressure (P1) = 100 atm
- Final pressure (P2) = 50 atm
- Ideal gas constant (R) = 2 cal/(K·mol)

### Step 2: Use the Formula for Entropy Change
For an isothermal process, the change in entropy (ΔS) can be calculated using the formula:
\[ \Delta S = nR \ln\left(\frac{P_1}{P_2}\right) \]

### Step 3: Substitute the Values into the Formula
Substituting the known values into the formula:
\[ \Delta S = 1 \, \text{mol} \times 2 \, \frac{\text{cal}}{\text{K}} \times \ln\left(\frac{100 \, \text{atm}}{50 \, \text{atm}}\right) \]

### Step 4: Simplify the Fraction Inside the Logarithm
Calculate the fraction:
\[ \frac{P_1}{P_2} = \frac{100}{50} = 2 \]

### Step 5: Calculate the Natural Logarithm
Now, calculate the natural logarithm of 2:
\[ \ln(2) \approx 0.693 \]

### Step 6: Substitute Back and Calculate ΔS
Now substitute this value back into the equation:
\[ \Delta S = 1 \times 2 \times 0.693 = 1.386 \, \text{cal/K} \]

### Step 7: Round the Result
Rounding to two decimal places gives:
\[ \Delta S \approx 1.39 \, \text{cal/K} \]

### Final Answer
The total entropy change in the process is approximately **1.39 cal/K**.

---