5 mole of an ideal gas of volume is expanded against vaccum to make its volume 2 times, then work done by the gas is:

Options

A. `-RT(V_2-V_1)`
B. `-RTln(V_2/V_1)`
C. 0
D. `C_v(T_2-T_1)`

Answer

To solve the problem, we need to determine the work done by an ideal gas when it expands against a vacuum. Let's break down the steps:

### Step 1: Understand the scenario
We have 5 moles of an ideal gas that is expanding against a vacuum. This means that there is no external pressure opposing the expansion of the gas.

### Step 2: Identify the initial and final volumes
Let the initial volume of the gas be \( V_1 \). According to the problem, the gas expands to a volume that is 2 times the initial volume, so the final volume \( V_2 \) will be:
\[ V_2 = 2V_1 \]

### Step 3: Recall the formula for work done
The work done \( W \) by the gas during expansion can be calculated using the formula:
\[ W = -P_{\text{external}} (V_2 - V_1) \]
where \( P_{\text{external}} \) is the external pressure.

### Step 4: Analyze the external pressure
Since the gas is expanding against a vacuum, the external pressure \( P_{\text{external}} \) is 0. Therefore, we can substitute this into the work done formula:
\[ W = -0 \times (V_2 - V_1) = 0 \]

### Step 5: Conclusion
Since the work done by the gas is calculated to be 0, we conclude that the work done by the gas during its expansion against a vacuum is:
\[ \text{Work done} = 0 \]

### Final Answer
The work done by the gas is **0**.

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