Calculate `DeltaS` for 3 mole of a diatomic ideal gas which is heated and compressed from 298 K and 1 bar to 596 K and 4 bar: [Given: `C_(v,m)(gas)=(5)/(2)R,"ln"(2)=0.70,R=2"cal K^(-1)mol^(-1)`]

Options

A. `-14.7"cal K"^(-1)`
B. `+14.7 cal K^(-1)`
C. `-4.9 cal K^(-1)`
D. 6.3 cal `K^(-1)`

Explanation

To calculate the change in entropy (ΔS) for 3 moles of a diatomic ideal gas heated and compressed from 298 K and 1 bar to 596 K and 4 bar, we can follow these steps:

### Step 1: Determine the heat capacity at constant pressure (Cp)
Given:
- \( C_{v,m} = \frac{5}{2} R \)

We know that:
\[ C_p = C_v + R \]
Substituting the value of \( C_v \):
\[ C_p = \frac{5}{2} R + R = \frac{5}{2} R + \frac{2}{2} R = \frac{7}{2} R \]

### Step 2: Substitute the value of R
Given:
- \( R = 2 \, \text{cal K}^{-1} \text{mol}^{-1} \)

Now substituting R into the equation for Cp:
\[ C_p = \frac{7}{2} \times 2 = 7 \, \text{cal K}^{-1} \text{mol}^{-1} \]

### Step 3: Use the formula for ΔS
The formula for the change in entropy (ΔS) for a system is given by:
\[ \Delta S = N C_p \ln\left(\frac{T_2}{T_1}\right) + N R \ln\left(\frac{P_1}{P_2}\right) \]
Where:
- \( N = 3 \, \text{moles} \)
- \( T_1 = 298 \, \text{K} \)
- \( T_2 = 596 \, \text{K} \)
- \( P_1 = 1 \, \text{bar} \)
- \( P_2 = 4 \, \text{bar} \)

### Step 4: Calculate the first term
Calculating the first term:
\[ \Delta S_1 = N C_p \ln\left(\frac{T_2}{T_1}\right) = 3 \times 7 \times \ln\left(\frac{596}{298}\right) \]
Calculating \( \frac{596}{298} = 2 \):
\[ \Delta S_1 = 3 \times 7 \times \ln(2) \]

### Step 5: Calculate the second term
Calculating the second term:
\[ \Delta S_2 = N R \ln\left(\frac{P_1}{P_2}\right) = 3 \times 2 \times \ln\left(\frac{1}{4}\right) = 3 \times 2 \times \ln(2^{-2}) = 3 \times 2 \times (-2 \ln(2)) \]
\[ \Delta S_2 = -12 \ln(2) \]

### Step 6: Combine the two terms
Now, combining both terms:
\[ \Delta S = \Delta S_1 + \Delta S_2 = (21 \ln(2)) + (-12 \ln(2)) = 9 \ln(2) \]

### Step 7: Substitute the value of ln(2)
Given:
- \( \ln(2) = 0.7 \)

Now substituting this value:
\[ \Delta S = 9 \times 0.7 = 6.3 \, \text{cal K}^{-1} \]

### Final Answer
Thus, the change in entropy \( \Delta S \) is:
\[ \Delta S = 6.3 \, \text{cal K}^{-1} \]

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