In an isothermal expansion change in internal energy and work done by the gas are respectively

Options

A. (-) ve, (-) ve
B. 0, (+) ve
C. 0, (-) ve
D. (+) ve, (-) ve

Answer

To solve the question regarding the change in internal energy and work done by the gas during an isothermal expansion, we can follow these steps:

### Step 1: Understand the Isothermal Process
In an isothermal process, the temperature of the system remains constant. This implies that the internal energy of an ideal gas, which is a function of temperature, does not change.

### Step 2: Determine the Change in Internal Energy
For an ideal gas, the change in internal energy (\( \Delta U \)) is given by:
\[ \Delta U = nC_V\Delta T \]
Where:
- \( n \) is the number of moles,
- \( C_V \) is the molar heat capacity at constant volume,
- \( \Delta T \) is the change in temperature.

Since the process is isothermal, \( \Delta T = 0 \). Thus:
\[ \Delta U = nC_V \cdot 0 = 0 \]
So, the change in internal energy is zero.

### Step 3: Determine the Work Done by the Gas
In an isothermal expansion, the work done (\( W \)) by the gas can be calculated using the formula:
\[ W = \int_{V_1}^{V_2} P \, dV \]
For an ideal gas, we can use the ideal gas law \( PV = nRT \) to express pressure \( P \) as:
\[ P = \frac{nRT}{V} \]
Thus, the work done during the expansion from volume \( V_1 \) to \( V_2 \) is:
\[ W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV \]
This integral evaluates to:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since \( V_2 > V_1 \) in an expansion, \( \ln\left(\frac{V_2}{V_1}\right) > 0 \). Therefore, the work done by the gas is positive.

### Conclusion
In an isothermal expansion:
- The change in internal energy (\( \Delta U \)) is **zero**.
- The work done by the gas (\( W \)) is **positive**.

Thus, the correct answer is:
**Change in internal energy: 0, Work done by the gas: Positive.**