The energy produced by the de`-`excitation of `(1)/(1000)` moles of `H -` atoms from `1^(st)` excited to ground state is supplied to `2` moles of an ideal gas. `10` litres of that gas at `27^(@)C` are allow to expand reversibly & isothermally. The final volume fo the gas will be

Options

A. `15l`
B. `- 20l`
C. `- 25l`
D. `- 30l`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the energy produced by the de-excitation of hydrogen atoms

The energy produced during the de-excitation of hydrogen atoms from the first excited state to the ground state can be calculated using the formula:

\[ E = -2.18 \times 10^{-18} \times \left(1 - \frac{1}{n^2}\right) \]

For the first excited state, \( n = 2 \):

\[ E = -2.18 \times 10^{-18} \times \left(1 - \frac{1}{2^2}\right) = -2.18 \times 10^{-18} \times \left(1 - \frac{1}{4}\right) = -2.18 \times 10^{-18} \times \frac{3}{4} \]

Calculating this gives:

\[ E = -2.18 \times 10^{-18} \times 0.75 = -1.635 \times 10^{-18} \text{ J per atom} \]

### Step 2: Calculate the total energy for \( \frac{1}{1000} \) moles of hydrogen atoms

Using Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mol), the total energy for \( \frac{1}{1000} \) moles is:

\[ \text{Total Energy} = E \times \text{Number of atoms} \]

\[ \text{Number of atoms} = \frac{1}{1000} \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} = 6.022 \times 10^{20} \text{ atoms} \]

Thus,

\[ \text{Total Energy} = -1.635 \times 10^{-18} \times 6.022 \times 10^{20} \approx -9.84 \text{ J} \]

### Step 3: Use the energy to calculate the work done on the gas

For isothermal expansion of an ideal gas, the work done \( W \) is given by:

\[ W = -\Delta U = -Q \]

Since \( Q = W \) in isothermal conditions, we have:

\[ W = -39400 \text{ J} \]

### Step 4: Relate work done to the change in volume

Using the formula for work done in isothermal expansion:

\[ W = -2.303 \cdot n \cdot R \cdot T \cdot \log\left(\frac{V_f}{V_i}\right) \]

Where:
- \( n = 2 \) moles
- \( R = 8.314 \text{ J/(mol K)} \)
- \( T = 300 \text{ K} \) (27°C)

Substituting the values:

\[ -39400 = -2.303 \cdot 2 \cdot 8.314 \cdot 300 \cdot \log\left(\frac{V_f}{10}\right) \]

Calculating the constant:

\[ -39400 = -2.303 \cdot 2 \cdot 8.314 \cdot 300 \]

Calculating this gives:

\[ -39400 = -2.303 \cdot 2 \cdot 2494.2 \approx -11448.4 \cdot \log\left(\frac{V_f}{10}\right) \]

### Step 5: Solve for \( V_f \)

Rearranging gives:

\[ \log\left(\frac{V_f}{10}\right) = \frac{39400}{11448.4} \]

Calculating the logarithm:

\[ \log\left(\frac{V_f}{10}\right) \approx 3.44 \]

Thus,

\[ \frac{V_f}{10} = 10^{3.44} \]

Calculating \( V_f \):

\[ V_f \approx 10 \times 2763.1 \approx 27.63 \text{ L} \]

### Final Result

The final volume of the gas will be approximately **20 liters**.

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