One mole of an ideal monoatomic gas expands reversibly and adiabatically from a volume of x litre to 14 litre at `27^(@)`C. Then value of x will be [Given, final temperature 189 K and `C_(V)` = 3/2 R].

Answer

To solve the problem step by step, we will use the concepts of thermodynamics related to adiabatic processes for an ideal monoatomic gas.

### Step 1: Identify the Given Values
- Initial volume, \( V_1 = x \) liters
- Final volume, \( V_2 = 14 \) liters
- Initial temperature, \( T_1 = 27^\circ C = 273 + 27 = 300 \, K \)
- Final temperature, \( T_2 = 189 \, K \)
- Heat capacity at constant volume, \( C_V = \frac{3}{2} R \)

### Step 2: Calculate the Value of \( \gamma \)
For a monoatomic ideal gas:
- \( C_P = C_V + R = \frac{3}{2} R + R = \frac{5}{2} R \)
- The ratio \( \gamma = \frac{C_P}{C_V} = \frac{\frac{5}{2} R}{\frac{3}{2} R} = \frac{5}{3} \)

### Step 3: Use the Adiabatic Condition
The adiabatic condition for an ideal gas states:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
This can be rearranged to:
\[ \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ \frac{300}{189} = \left(\frac{14}{x}\right)^{\frac{5}{3} - 1} \]
This simplifies to:
\[ \frac{300}{189} = \left(\frac{14}{x}\right)^{\frac{2}{3}} \]

### Step 5: Raise Both Sides to the Power of \( \frac{3}{2} \)
To eliminate the exponent, raise both sides to the power of \( \frac{3}{2} \):
\[ \left(\frac{300}{189}\right)^{\frac{3}{2}} = \frac{14^2}{x^2} \]

### Step 6: Calculate \( \frac{300}{189} \)
Calculating \( \frac{300}{189} \):
\[ \frac{300}{189} \approx 1.5873 \]
Now raise it to the power of \( \frac{3}{2} \):
\[ \left(1.5873\right)^{\frac{3}{2}} \approx 1.5873^{1.5} \approx 1.947 \]

### Step 7: Set Up the Equation
Now we have:
\[ 1.947 = \frac{196}{x^2} \]
Where \( 14^2 = 196 \).

### Step 8: Solve for \( x^2 \)
Rearranging gives:
\[ x^2 = \frac{196}{1.947} \approx 100.5 \]

### Step 9: Calculate \( x \)
Taking the square root:
\[ x \approx \sqrt{100.5} \approx 10.03 \, \text{liters} \]

### Step 10: Round to the Nearest Value
Since we need a practical answer, we can round \( x \) to:
\[ x \approx 10 \, \text{liters} \]

### Final Answer
The initial volume \( x \) is approximately **10 liters**.

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