If one mole of gas doubles its volume at temperature T isothermally then work done by the gas is

Options

A. `-RT log_e 4`
B. `RT log_e 2`
C. `-RT log_(10) 2`
D. RT

Answer

To solve the problem of calculating the work done by one mole of gas that doubles its volume at a constant temperature (isothermally), we can use the formula for work done during an isothermal expansion of an ideal gas. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final Volumes
Let the initial volume of the gas be \( V_1 = V \) and the final volume after doubling be \( V_2 = 2V \).

### Step 2: Use the Formula for Work Done in Isothermal Process
The work done \( W \) by an ideal gas during an isothermal expansion can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
where:
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( V_2 \) is the final volume,
- \( V_1 \) is the initial volume.

### Step 3: Substitute the Values
Since we have one mole of gas (\( n = 1 \)), we can substitute \( V_1 \) and \( V_2 \) into the formula:
\[ W = 1 \cdot R \cdot T \cdot \ln\left(\frac{2V}{V}\right) \]

### Step 4: Simplify the Expression
The logarithmic term simplifies as follows:
\[ \ln\left(\frac{2V}{V}\right) = \ln(2) \]
Thus, the work done becomes:
\[ W = RT \ln(2) \]

### Final Answer
Therefore, the work done by the gas when it doubles its volume is:
\[ W = RT \ln(2) \]

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