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. Here are the steps to find the total internal energy (U):

### Step 1: Identify the given data
- Volume of liquid \( V_1 = 100 \) liters
- Change in volume \( \Delta V = 2 \) liters
- External pressure \( P_{ext} = 1 \) atm
- Heat supplied \( Q = 202.6 \) joules
- Change in enthalpy \( \Delta H = Q = 202.6 \) joules

### Step 2: Calculate the work done (W)
The work done on/by the system can be calculated using the formula:
\[ W = -P_{ext} \Delta V \]
First, we need to convert the volume change from liters to cubic meters (1 L = 0.001 m³):
\[ \Delta V = 2 \, \text{L} = 2 \times 0.001 \, \text{m}^3 = 0.002 \, \text{m}^3 \]
Now, convert the external pressure from atm to pascals (1 atm = 101.325 kPa = 101325 Pa):
\[ P_{ext} = 1 \, \text{atm} = 101325 \, \text{Pa} \]
Now substitute the values into the work done formula:
\[ W = -101325 \, \text{Pa} \times 0.002 \, \text{m}^3 = -202.65 \, \text{J} \]

### Step 3: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q + W \]
Substituting the values we have:
\[ \Delta U = 202.6 \, \text{J} + (-202.65 \, \text{J}) = 202.6 \, \text{J} - 202.65 \, \text{J} = -0.05 \, \text{J} \]

### Step 4: Conclusion
The total internal energy change \( \Delta U \) is:
\[ \Delta U = -0.05 \, \text{J} \]