The relation of internal energy, enthalpy and work done can be represented by:

Options

A. `DeltaE= DeltaH+W`
B. `DeltaE = W- DeltaH`
C. `DeltaH= DeltaE+W`
D. `W= DeltaE- DeltaH`

Explanation

To derive the relationship between internal energy (ΔU), enthalpy (ΔH), and work done (W), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Enthalpy**:
Enthalpy (H) is defined as the sum of the internal energy (U) of a system and the product of its pressure (P) and volume (V):
\[ H = U + PV \]

2. **Change in Enthalpy**:
The change in enthalpy (ΔH) can be expressed as:
\[ \Delta H = \Delta U + P\Delta V \]
Here, ΔU is the change in internal energy, and PΔV is the work done when the pressure is constant.

3. **First Law of Thermodynamics**:
According to the first law of thermodynamics, the change in internal energy (ΔU) is given by:
\[ \Delta U = Q + W \]
where Q is the heat added to the system and W is the work done by the system.

4. **Substituting Work Done**:
In a constant pressure process, the work done (W) can be expressed as:
\[ W = -P\Delta V \]
Substituting this into the equation for ΔU gives:
\[ \Delta U = Q - P\Delta V \]

5. **Combining Equations**:
Now, we can substitute ΔU from the first law into the equation for ΔH:
\[ \Delta H = (Q - P\Delta V) + P\Delta V \]
Simplifying this, we find:
\[ \Delta H = Q \]
when the pressure is constant.

6. **Final Relationship**:
Rearranging the equation for ΔH gives us the relationship between ΔU, ΔH, and W:
\[ \Delta U = \Delta H - P\Delta V \]
Since \(W = -P\Delta V\), we can express this as:
\[ \Delta U = \Delta H + W \]

### Final Result:
Thus, the relationship between internal energy, enthalpy, and work done can be summarized as:
\[ \Delta U = \Delta H + W \]