One mole of component `P` and two moles of component `Q` are mixed at `27^(@)C` to form an ideal binary solution. Select the correct option for given information?

Options

A. a)
`DeltaS_("mix")=3.82"cal"k^(-1)`
B. b) `DeltaH_("mix")=0`
C. c) `DeltaS_("mix")=0`
D. d) `DeltaG_("mix")=-1145.7` cal

Explanation

To solve the problem of mixing one mole of component P and two moles of component Q at 27°C to form an ideal binary solution, we will follow these steps:

### Step 1: Identify the components and their moles
- Let component P be the solute with \(N_P = 1\) mole.
- Let component Q be the solvent with \(N_Q = 2\) moles.

### Step 2: Calculate the total number of moles
Total moles \(N_{total} = N_P + N_Q = 1 + 2 = 3\) moles.

### Step 3: Calculate the mole fractions
- Mole fraction of P (\(X_P\)):
\[ X_P = \frac{N_P}{N_{total}} = \frac{1}{3} \]

- Mole fraction of Q (\(X_Q\)):
\[ X_Q = \frac{N_Q}{N_{total}} = \frac{2}{3} \]

### Step 4: Use the Gibbs free energy of mixing formula
For an ideal binary solution, the Gibbs free energy of mixing (\(\Delta G_{mix}\)) is given by:
\[ \Delta G_{mix} = RT \left( N_P \ln X_P + N_Q \ln X_Q \right) \]
Where:
- \(R\) is the universal gas constant (approximately \(2 \, \text{cal/K·mol}\)),
- \(T\) is the temperature in Kelvin (27°C = 300 K).

### Step 5: Substitute the values into the Gibbs free energy equation
\[ \Delta G_{mix} = 2 \, \text{cal/K·mol} \times 300 \, \text{K} \left( 1 \ln \left( \frac{1}{3} \right) + 2 \ln \left( \frac{2}{3} \right) \right) \]

### Step 6: Calculate the logarithmic values
- \(\ln \left( \frac{1}{3} \right) \approx -1.0986\)
- \(\ln \left( \frac{2}{3} \right) \approx -0.4055\)

### Step 7: Plug in the logarithmic values
\[ \Delta G_{mix} = 2 \times 300 \left( 1 \times (-1.0986) + 2 \times (-0.4055) \right) \]
\[ = 600 \left( -1.0986 - 0.8110 \right) \]
\[ = 600 \times (-1.9096) \approx -1145.76 \, \text{calories} \]

### Step 8: Determine the change in enthalpy and entropy
For an ideal solution:
- \(\Delta H_{mix} = 0\) (no heat absorbed or released during mixing).
- Use the relation:
\[ \Delta G_{mix} = \Delta H_{mix} - T \Delta S_{mix} \]
Since \(\Delta H_{mix} = 0\):
\[ \Delta G_{mix} = -T \Delta S_{mix} \implies \Delta S_{mix} = -\frac{\Delta G_{mix}}{T} \]
Substituting the values:
\[ \Delta S_{mix} = -\frac{-1145.76}{300} \approx 3.82 \, \text{cal/K} \]

### Conclusion
The results are:
- \(\Delta G_{mix} \approx -1145.76 \, \text{calories}\)
- \(\Delta H_{mix} = 0\)
- \(\Delta S_{mix} \approx 3.82 \, \text{cal/K}\)

### Correct Options
Based on the calculations:
- Option A: \(\Delta S_{mix} = 3.82 \, \text{cal/K}\) (Correct)
- Option B: \(\Delta H_{mix} = 0\) (Correct)
- Option D: \(\Delta G_{mix} = -1145.7 \, \text{calories}\) (Correct)