If `DeltaU` and `Delta W` represent the increase in internal energy and work done by the system resectively in a thermodynamical process, which of the following is true?

Options

A. `Delta U = - Delta W`, in an adiabatic process
B. `Delta U = Delta W`, in an isothermal process
C. `Delta U = Delta W`, in an adiabatic process
D. `Delta U = - Delta W`, in an isothermal process

Answer

To solve the question regarding the relationship between the change in internal energy (ΔU) and the work done by the system (ΔW) in a thermodynamic process, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First Law of Thermodynamics**:
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (ΔQ) minus the work done by the system (ΔW). This can be expressed mathematically as:
\[ \Delta Q = \Delta U + \Delta W \]

2. **Consider Different Thermodynamic Processes**:
- **Adiabatic Process**: In an adiabatic process, there is no heat exchange with the surroundings, which means ΔQ = 0. Therefore, from the first law:
\[ 0 = \Delta U + \Delta W \implies \Delta U = -\Delta W \]
- **Isothermal Process**: In an isothermal process for an ideal gas, the internal energy remains constant (ΔU = 0). Thus:
\[ 0 = \Delta Q + \Delta W \implies \Delta Q = -\Delta W \]

3. **Analyze the Signs of ΔU and ΔW**:
- When the system does work on the surroundings, ΔW is considered negative (since work is done by the system). Therefore, if ΔU = -ΔW, it implies that an increase in internal energy corresponds to a decrease in work done by the system.
- In an adiabatic process, if the internal energy increases, the work done by the system decreases and vice versa.

4. **Conclusion**:
Based on the analysis above, the correct relationship for an adiabatic process is:
\[ \Delta U = -\Delta W \]

### Final Answer:
The correct statement is that in an adiabatic process, ΔU = -ΔW.