A vessel contains `100` litres of a liquid `X`. Heat is supplied to the liquid in such a fashion that, Heat given `=` change in enthalpy. The volume of the liquid increases by `2` litres. If the external pressure is one atm, and `202.6` joules of heat were supplied then, [`U`-total internal energy]

Options

A. `Delta U=0, Delta H=0`
B. `Delta U+202.6J, Delta H= +202.6 J, Delta H = +202.6 J`
C. `Delta U= -202.6J, Delta H= -202.6 J`
D. `Delta U=0, Delta H+202.6J`

Answer

To solve the problem, we will use the first law of thermodynamics and the relationship between enthalpy, internal energy, and work done.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Volume of liquid \( X = 100 \) liters
- Change in volume \( \Delta V = 2 \) liters
- External pressure \( P_{\text{external}} = 1 \) atm
- Heat supplied \( Q = 202.6 \) joules

2. **Convert External Pressure to Joules:**
- We need to convert the external pressure from atm to joules. The conversion factor is:
\[ 1 \text{ atm} = 101.325 \text{ J/L} \]
- Therefore, the work done \( W \) can be calculated as:
\[ W = -P_{\text{external}} \times \Delta V = -1 \text{ atm} \times 2 \text{ L} = -2 \text{ L atm} \]
- Now convert \( -2 \text{ L atm} \) to joules:
\[ W = -2 \text{ L atm} \times 101.325 \text{ J/L atm} = -202.65 \text{ J} \]

3. **Apply the First Law of Thermodynamics:**
- The first law of thermodynamics is given by:
\[ \Delta U = Q + W \]
- Substitute the values of \( Q \) and \( W \):
\[ \Delta U = 202.6 \text{ J} + (-202.65 \text{ J}) = 202.6 \text{ J} - 202.65 \text{ J} = -0.05 \text{ J} \]
- Since we are looking for the change in internal energy, we can round this to:
\[ \Delta U \approx 0 \text{ J} \]

4. **Determine Change in Enthalpy:**
- According to the problem, the heat given is equal to the change in enthalpy:
\[ \Delta H = Q = 202.6 \text{ J} \]

5. **Final Results:**
- Change in internal energy \( \Delta U \approx 0 \text{ J} \)
- Change in enthalpy \( \Delta H = 202.6 \text{ J} \)

### Conclusion:
The final answers are:
- \( \Delta U = 0 \text{ J} \)
- \( \Delta H = 202.6 \text{ J} \)