When 1 mol of `CO_(2)(g)` occupying a volumeof 10 L at `27^(@)` isallowed to expand under adiabatic conditions, temperature fallsto `-123^(@)C`. Hence, final volume of the gas will be `:`

Options

A. 20L
B. 40L
C. 60L
D. 80L

Explanation

To solve the problem, we will use the adiabatic process equation for an ideal gas, which relates the temperatures and volumes before and after the expansion. The equation is given by:

\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]

Where:
- \(T_1\) = initial temperature in Kelvin
- \(T_2\) = final temperature in Kelvin
- \(V_1\) = initial volume
- \(V_2\) = final volume
- \(\gamma\) = heat capacity ratio (Cp/Cv)

### Step 1: Convert temperatures to Kelvin
- Initial temperature \(T_1 = 27^\circ C = 27 + 273 = 300 \, K\)
- Final temperature \(T_2 = -123^\circ C = -123 + 273 = 150 \, K\)

### Step 2: Identify the initial volume
- Given \(V_1 = 10 \, L\)

### Step 3: Determine the value of \(\gamma\) for \(CO_2\)
For carbon dioxide (\(CO_2\)), the value of \(\gamma\) is approximately \(1.3\).

### Step 4: Substitute the values into the adiabatic equation
We can rearrange the adiabatic equation to solve for \(V_2\):

\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]

Rearranging gives:

\[ \frac{V_1}{V_2} = \left(\frac{T_2}{T_1}\right)^{\frac{1}{\gamma - 1}} \]

Now substituting the known values:

\[ \frac{V_1}{V_2} = \left(\frac{150}{300}\right)^{\frac{1}{1.3 - 1}} = \left(\frac{1}{2}\right)^{\frac{1}{0.3}} = \left(\frac{1}{2}\right)^{\frac{10}{3}} = \left(\frac{1}{2}\right)^{3.33} \]

Calculating \(\left(\frac{1}{2}\right)^{3.33}\):

\[ \left(\frac{1}{2}\right)^{3.33} \approx 0.096 \]

### Step 5: Solve for \(V_2\)
Now we can find \(V_2\):

\[ \frac{10}{V_2} = 0.096 \implies V_2 = \frac{10}{0.096} \approx 104.17 \, L \]

### Final Answer
The final volume \(V_2\) of the gas after adiabatic expansion is approximately \(104.17 \, L\).

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