An ideal gas system undergoes an isothermal process, then the work done during the process is

Options

A. `nRTIn(V_(2))/(V_(1))`
B. `nRTIn(V_(1))/(V_(2))`
C. `2nRTIn(V_(2))/(V_(1))`
D. `2nRTIn(V_(1))/(V_(2))`

Explanation

To find the work done during an isothermal process for an ideal gas, we can follow these steps:

### Step 1: Understand the Isothermal Process
In an isothermal process, the temperature (T) remains constant. For an ideal gas, this means that the product of pressure (P) and volume (V) is constant, given by the equation:
\[ PV = nRT \]
where \( n \) is the number of moles of gas and \( R \) is the universal gas constant.

### Step 2: Express Work Done
The infinitesimal work done (dW) during a small change in volume (dV) can be expressed as:
\[ dW = P \, dV \]
To find the total work done (W), we need to integrate this expression over the volume change from \( V_1 \) to \( V_2 \):
\[ W = \int_{V_1}^{V_2} P \, dV \]

### Step 3: Substitute for Pressure
Since we know that \( P = \frac{nRT}{V} \) for an ideal gas during an isothermal process, we can substitute this into our work equation:
\[ W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV \]

### Step 4: Perform the Integration
Now we can perform the integration:
\[ W = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV \]
The integral of \( \frac{1}{V} \) is:
\[ \int \frac{1}{V} \, dV = \ln V \]
Thus, we have:
\[ W = nRT [\ln V]_{V_1}^{V_2} \]
This evaluates to:
\[ W = nRT (\ln V_2 - \ln V_1) \]

### Step 5: Simplify the Expression
Using the properties of logarithms, we can simplify this to:
\[ W = nRT \ln \left(\frac{V_2}{V_1}\right) \]

### Final Result
Therefore, the work done during an isothermal process for an ideal gas is given by:
\[ W = nRT \ln \left(\frac{V_2}{V_1}\right) \]