Ideal gas 1 mole expand isothermally reversibly 2 lt. to 4lt and same gas 3 mole expand from 2 lt. to x lt and doing same work, what is 'x'

Options

A. `(8)^(1/3)`
B. `(4)^(2/3)`
C. 2
D. 4 lt

Explanation

To solve the problem, we need to use the formula for work done during isothermal reversible expansion of an ideal gas. The work done (W) can be expressed as:

\[ W = nRT \ln \left( \frac{V_f}{V_i} \right) \]

where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = absolute temperature
- \( V_f \) = final volume
- \( V_i \) = initial volume

### Step 1: Calculate Work Done by the First Gas
For the first gas (1 mole expanding from 2 L to 4 L):

- \( n_1 = 1 \) mole
- \( V_i = 2 \) L
- \( V_f = 4 \) L

Using the formula:

\[ W_1 = n_1RT \ln \left( \frac{V_f}{V_i} \right) \]
\[ W_1 = 1RT \ln \left( \frac{4}{2} \right) \]
\[ W_1 = 1RT \ln(2) \]

### Step 2: Calculate Work Done by the Second Gas
For the second gas (3 moles expanding from 2 L to \( x \) L):

- \( n_2 = 3 \) moles
- \( V_i = 2 \) L
- \( V_f = x \) L

Using the formula:

\[ W_2 = n_2RT \ln \left( \frac{V_f}{V_i} \right) \]
\[ W_2 = 3RT \ln \left( \frac{x}{2} \right) \]

### Step 3: Set the Work Done Equal
Since both expansions do the same work:

\[ W_1 = W_2 \]
\[ 1RT \ln(2) = 3RT \ln \left( \frac{x}{2} \right) \]

### Step 4: Simplify the Equation
We can cancel \( RT \) from both sides (assuming \( R \) and \( T \) are not zero):

\[ \ln(2) = 3 \ln \left( \frac{x}{2} \right) \]

### Step 5: Use Properties of Logarithms
Using the property of logarithms, we can rewrite the right side:

\[ \ln(2) = \ln \left( \left( \frac{x}{2} \right)^3 \right) \]

### Step 6: Exponentiate Both Sides
Exponentiating both sides gives:

\[ 2 = \left( \frac{x}{2} \right)^3 \]

### Step 7: Solve for \( x \)
Now we can solve for \( x \):

\[ 2 = \frac{x^3}{8} \]
\[ x^3 = 16 \]
\[ x = 16^{1/3} \]
\[ x = 4 \]

### Conclusion
The final volume \( x \) is 4 L.