As per second law of thermodynamics a process taken place spontaneously if and only if the entropy of the universe increases due to the process. Change in entropy is given by
`Delta S = (Q_(rev))/(T)`
A gas `C_(V) = (0.2 T) Cal K^(-1)`. What is the change in its entropy when one mole of it is heated from `27^(@)C` to `127^(@)C` at constant volume ?

Options

A. `20calK^(-1) mol^(-1)`
B. `15calK^(-1) mol^(-1)`
C. `35calK^(-1) mol^(-1)`
D. `25calK^(-1) mol^(-1)`

Answer

To find the change in entropy (\(\Delta S\)) of one mole of a gas with a specific heat capacity at constant volume (\(C_V\)) of \(0.2 T\) cal K\(^{-1}\) when heated from \(27^\circ C\) to \(127^\circ C\) at constant volume, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
- The initial temperature \(T_1\) is \(27^\circ C\).
- The final temperature \(T_2\) is \(127^\circ C\).
- Convert these temperatures to Kelvin:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
\[ T_2 = 127 + 273 = 400 \, \text{K} \]

### Step 2: Write the formula for change in entropy at constant volume
For a process at constant volume, the change in entropy (\(\Delta S\)) can be calculated using the formula:
\[ \Delta S = n C_V \int_{T_1}^{T_2} \frac{dT}{T} \]
Where:
- \(n\) = number of moles (1 mole in this case)
- \(C_V\) = \(0.2 T\) cal K\(^{-1}\)

### Step 3: Substitute \(C_V\) into the entropy formula
Substituting \(C_V\) into the equation:
\[ \Delta S = 1 \cdot \int_{300}^{400} 0.2 T \frac{dT}{T} \]
This simplifies to:
\[ \Delta S = 0.2 \int_{300}^{400} dT \]

### Step 4: Evaluate the integral
The integral \(\int_{300}^{400} dT\) is simply:
\[ \Delta S = 0.2 \cdot (400 - 300) \]
\[ \Delta S = 0.2 \cdot 100 = 20 \, \text{cal K}^{-1} \]

### Step 5: Conclusion
Thus, the change in entropy when one mole of the gas is heated from \(27^\circ C\) to \(127^\circ C\) at constant volume is:
\[ \Delta S = 20 \, \text{cal K}^{-1} \]

### Final Answer:
The change in entropy is \(20 \, \text{cal K}^{-1}\).

---