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

Options

A. `nRTln((V_(2))/(V_(1)))`
B. `nRTln((V_(1))/(V_(2)))`
C. `2nRTln((V_(2))/(V_(1)))`
D. `2nRTln((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
An isothermal process is one in which the temperature of the system remains constant. For an ideal gas, this means that the product of pressure (P) and volume (V) is constant (PV = constant).

### Step 2: Write the Expression for Work Done
The work done (W) during a process can be expressed as the integral of pressure with respect to volume:
\[ W = \int_{V_1}^{V_2} P \, dV \]

### Step 3: Use the Ideal Gas Law
From the ideal gas law, we know:
\[ PV = nRT \]
Where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = absolute temperature (constant for isothermal process)

Since \( PV = nRT \) is constant, we can express pressure (P) in terms of volume (V):
\[ P = \frac{nRT}{V} \]

### Step 4: Substitute Pressure into the Work Integral
Substituting the expression for pressure into the work integral gives:
\[ W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV \]

### Step 5: Integrate
The integral of \( \frac{1}{V} \) is:
\[ \int \frac{1}{V} \, dV = \ln V \]
Thus, we have:
\[ W = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV = nRT [\ln V]_{V_1}^{V_2} \]

### Step 6: Apply the Limits of Integration
Applying the limits \( V_1 \) and \( V_2 \):
\[ W = nRT (\ln V_2 - \ln V_1) \]
Using the properties of logarithms, this can be simplified to:
\[ W = nRT \ln \left(\frac{V_2}{V_1}\right) \]

### Final Result
The work done during the isothermal process is:
\[ W = nRT \ln \left(\frac{V_2}{V_1}\right) \]