If heat is supplied to an ideal gas in an isothermal process.

Options

A. the internal energy of the gas will increase
B. the gas will do positive work
C. the gas will do negative work
D. the said process is not possible.

Explanation

To solve the problem of what happens when heat is supplied to an ideal gas in an isothermal process, we can follow these steps:

### Step 1: Understand the Isothermal Process
In an isothermal process, the temperature of the system remains constant. For an ideal gas, this means that any heat added to the system does not change the internal energy of the gas.

**Hint:** Remember that in an isothermal process, the temperature (T) is constant throughout the process.

### Step 2: Relate Internal Energy to Temperature
The internal energy (U) of an ideal gas is a function of temperature. For an ideal gas, the change in internal energy (ΔU) can be expressed as:
\[ \Delta U = nC_v \Delta T \]
where \(n\) is the number of moles, \(C_v\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature.

**Hint:** Since the process is isothermal, think about what happens to \(\Delta T\).

### Step 3: Determine the Change in Internal Energy
Since the temperature is constant in an isothermal process, we have:
\[ \Delta T = 0 \]
Thus, substituting into the equation for internal energy:
\[ \Delta U = nC_v \cdot 0 = 0 \]
This indicates that there is no change in internal energy when heat is supplied.

**Hint:** Consider what it means for internal energy to not change when heat is added.

### Step 4: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta Q = \Delta W + \Delta U \]
Since we found that \(\Delta U = 0\), the equation simplifies to:
\[ \Delta Q = \Delta W \]
This means that the heat added to the system is entirely converted into work done by the gas.

**Hint:** Think about the implications of heat being converted entirely into work.

### Step 5: Determine the Work Done by the Gas
Since heat is being supplied to the gas (which means \(\Delta Q > 0\)), and it is equal to the work done by the gas, we conclude that:
\[ \Delta W > 0 \]
This indicates that the gas does positive work on the surroundings.

**Hint:** Reflect on what it means for the gas to do positive work in terms of its expansion.

### Conclusion
From the analysis above, we can conclude:
- The internal energy of the gas does not increase (it remains constant).
- The gas does positive work when heat is supplied.

Thus, the correct answer to the question is that the gas will do positive work.

### Final Answer
The gas will do a positive work.