In an isothermal process

Options

A. q = 0 and `Delta U` = 0
B. q `ne` 0 and `Delta U` = 0
C. q = 0 and `Delta U ne` 0
D. `q ne 0 and Delta U ne 0`

Answer

To solve the problem regarding an isothermal process, we will follow these steps:

### Step 1: Understand the Isothermal Process
In an isothermal process, the temperature (T) of the system remains constant. This implies that the change in temperature (ΔT) is equal to zero.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
Where:
- \( Q \) = heat added to the system
- \( \Delta U \) = change in internal energy
- \( W \) = work done by the system

### Step 3: Determine the Change in Internal Energy (ΔU)
For an ideal gas, the change in internal energy (ΔU) is related to the change in temperature (ΔT) by the equation:
\[ \Delta U = n C_V \Delta T \]
Since we established that ΔT = 0 for an isothermal process, we can substitute this into the equation:
\[ \Delta U = n C_V (0) = 0 \]
Thus, we find:
\[ \Delta U = 0 \]

### Step 4: Substitute ΔU into the First Law Equation
Now that we have ΔU, we can substitute it back into the first law equation:
\[ Q = \Delta U + W \]
This becomes:
\[ Q = 0 + W \]
Thus, we conclude:
\[ Q = W \]

### Step 5: Analyze the Results
Since \( Q \) is equal to \( W \) and we know that in an isothermal process, heat can be transferred to or from the system, \( Q \) is not equal to zero. Therefore, we have:
- \( Q \neq 0 \)
- \( \Delta U = 0 \)

### Conclusion
In summary, for an isothermal process:
- Heat added to the system (Q) is not zero.
- Change in internal energy (ΔU) is zero.