### Step-by-Step Solution
**Step 1: State the First Law of Thermodynamics**
The First Law of Thermodynamics states that the change in internal energy (ΔU) of a closed system is equal to the heat added to the system (Q) minus the work done by the system (W). Mathematically, it can be expressed as:
\[ \Delta U = Q - W \]
This law essentially reflects the principle of conservation of energy, indicating that energy cannot be created or destroyed, only transformed from one form to another.
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**Step 2: Explanation of the First Law**
In a thermodynamic process, when heat is added to a system, it can either increase the internal energy of the system or do work on the surroundings. If the system does work, then some of the energy goes into doing that work, and the remaining energy contributes to the internal energy of the system. Conversely, if work is done on the system, it can increase the internal energy.
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**Step 3: Application of the First Law to Isothermal Process**
In an isothermal process, the temperature of the system remains constant. For an ideal gas, the internal energy (U) is a function of temperature. Therefore, if the temperature is constant, the change in internal energy (ΔU) is zero:
\[ \Delta U = 0 \]
Applying the First Law of Thermodynamics:
\[ Q - W = \Delta U \]
Substituting ΔU = 0:
\[ Q = W \]
This means that the heat added to the system is entirely converted into work done by the system. If heat is absorbed (Q > 0), the system does work on the surroundings (W > 0). Conversely, if the system releases heat (Q < 0), work is done on the system (W < 0).
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**Step 4: Application of the First Law to Adiabatic Process**
In an adiabatic process, there is no heat exchange with the surroundings (Q = 0). Therefore, applying the First Law of Thermodynamics:
\[ \Delta U = Q - W \]
Substituting Q = 0:
\[ \Delta U = -W \]
This indicates that any work done by the system results in a decrease in internal energy. If the system does work on the surroundings (W > 0), the internal energy decreases (ΔU < 0), which typically results in a drop in temperature. Conversely, if work is done on the system (W < 0), the internal energy increases (ΔU > 0), leading to a rise in temperature.
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