Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation.
`DeltaS=2.303 "nC"_(v)"log"((T_(2))/(T_(1))) + 2.303 "nR log" ((V_(2))/(V_(1)))`
`DeltaS=2.303 "nC"_(P) "log"((T_(2))/(T_(1))) + 2.303 "nR log" ((P_(1))/(P_(2)))`
Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T`
An isobaric process having one mole of ideal gas has entropy change `23.03` J/K for the temperature range `27^(@)` C to `327^(@)` C . What would be the molar specific heat capacity `(C_(v))`?

Options

A. `(10)/("log"2) J//K mol`
B. `(10)/("log"2) - 8.3 J//K` mol
C. `10 xx "log"2 J//K mol`
D. `10 log2 + 8.3 J//k` mol

Explanation

To solve the problem, we need to determine the molar specific heat capacity \( C_v \) for an ideal gas given the entropy change \( \Delta S \) during an isobaric process. We will use the provided equation for entropy change and the relationship between \( C_p \) and \( C_v \).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Entropy change, \( \Delta S = 23.03 \, \text{J/K} \)
- Initial temperature, \( T_1 = 27^\circ C = 300 \, \text{K} \)
- Final temperature, \( T_2 = 327^\circ C = 600 \, \text{K} \)
- Number of moles, \( n = 1 \)

2. **Use the Entropy Change Equation**:
For an isobaric process, the entropy change can be expressed as:
\[ \Delta S = 2.303 \cdot n \cdot C_p \cdot \log\left(\frac{T_2}{T_1}\right) \]
Substituting the known values:
\[ 23.03 = 2.303 \cdot 1 \cdot C_p \cdot \log\left(\frac{600}{300}\right) \]

3. **Calculate the Logarithm**:
\[ \log\left(\frac{600}{300}\right) = \log(2) \approx 0.301 \]

4. **Substitute Logarithm Value**:
Now substitute the logarithm value back into the equation:
\[ 23.03 = 2.303 \cdot C_p \cdot 0.301 \]

5. **Solve for \( C_p \)**:
Rearranging the equation to solve for \( C_p \):
\[ C_p = \frac{23.03}{2.303 \cdot 0.301} \]
Calculate \( C_p \):
\[ C_p \approx \frac{23.03}{0.693} \approx 33.23 \, \text{J/K} \]

6. **Use the Relationship Between \( C_p \) and \( C_v \)**:
The relationship between \( C_p \) and \( C_v \) for an ideal gas is given by:
\[ C_p - C_v = R \]
Where \( R = 8.314 \, \text{J/(mol K)} \).

7. **Calculate \( C_v \)**:
Rearranging gives:
\[ C_v = C_p - R \]
Substituting the values:
\[ C_v = 33.23 - 8.314 \approx 24.92 \, \text{J/K} \]

### Final Answer:
The molar specific heat capacity \( C_v \) is approximately \( 24.92 \, \text{J/K} \).