One mole of ideal monatomic gas was taken through isochoric heating from `100` K to `1000` K. Calculate `DeltaS_("system"), Delta_("surr")` and `DeltaS_("total")` in
`(i)` When the process carried out reversibly `" "(ii)` When the process carried out irreversibly (one step)

Explanation

To solve the problem, we need to calculate the change in entropy for the system, the surroundings, and the total entropy for both reversible and irreversible processes during isochoric heating of one mole of an ideal monatomic gas from 100 K to 1000 K.

### Step-by-Step Solution

### Given:
- Number of moles, \( n = 1 \)
- Initial temperature, \( T_1 = 100 \, K \)
- Final temperature, \( T_2 = 1000 \, K \)
- For a monatomic ideal gas, the molar heat capacity at constant volume, \( C_V = \frac{3}{2} R \)

### (i) Reversible Process

1. **Calculate the change in entropy of the system (\( \Delta S_{\text{system}} \))**:
\[ \Delta S_{\text{system}} = \int_{T_1}^{T_2} \frac{dQ}{T} \]
Since the process is isochoric, \( dQ = n C_V dT \):
\[ \Delta S_{\text{system}} = \int_{T_1}^{T_2} \frac{n C_V dT}{T} \]
Substituting \( C_V = \frac{3}{2} R \):
\[ \Delta S_{\text{system}} = n \cdot \frac{3}{2} R \int_{T_1}^{T_2} \frac{dT}{T} \]
\[ = \frac{3}{2} R \ln\left(\frac{T_2}{T_1}\right) \]
\[ = \frac{3}{2} R \ln\left(\frac{1000}{100}\right) = \frac{3}{2} R \ln(10) \]

2. **Calculate the change in entropy of the surroundings (\( \Delta S_{\text{surr}} \))**:
\[ \Delta S_{\text{surr}} = -\frac{dQ}{T_{\text{surr}}} \]
For isochoric heating, we can use the final temperature of the system as the temperature of the surroundings:
\[ dQ = n C_V (T_2 - T_1) = n \cdot \frac{3}{2} R (1000 - 100) \]
\[ = \frac{3}{2} R (900) \]
Thus,
\[ \Delta S_{\text{surr}} = -\frac{\frac{3}{2} R (900)}{1000} \]
\[ = -\frac{3}{2} R \cdot 0.9 = -\frac{27}{20} R \]

3. **Calculate the total change in entropy (\( \Delta S_{\text{total}} \))**:
\[ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surr}} \]
\[ = \frac{3}{2} R \ln(10) - \frac{27}{20} R \]

### (ii) Irreversible Process

1. **Calculate the change in entropy of the system (\( \Delta S_{\text{system}} \))**:
The change in entropy for the system remains the same as in the reversible case:
\[ \Delta S_{\text{system}} = \frac{3}{2} R \ln(10) \]

2. **Calculate the change in entropy of the surroundings (\( \Delta S_{\text{surr}} \))**:
The calculation for the surroundings remains the same:
\[ \Delta S_{\text{surr}} = -\frac{27}{20} R \]

3. **Calculate the total change in entropy (\( \Delta S_{\text{total}} \))**:
The total change in entropy is still:
\[ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surr}} \]
\[ = \frac{3}{2} R \ln(10) - \frac{27}{20} R \]

### Final Results
- \( \Delta S_{\text{system}} = \frac{3}{2} R \ln(10) \)
- \( \Delta S_{\text{surr}} = -\frac{27}{20} R \)
- \( \Delta S_{\text{total}} = \frac{3}{2} R \ln(10) - \frac{27}{20} R \)