a monoatomic gas `(Cv=(3)/(2)R)` is allowed to expand adiabaticaly and reversibly from initial volume of 8 L to 300 K to a volume `V_(2)` at 250 K. `V_(2)` is :

Options

A. 10.5 L
B. 23 L
C. 8.5 L
D. 50.5 L

Answer

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

### Step-by-Step Solution:

**Step 1: Identify Given Values**
- Initial volume \( V_1 = 8 \, \text{L} \)
- Initial temperature \( T_1 = 300 \, \text{K} \)
- Final temperature \( T_2 = 250 \, \text{K} \)
- For a monoatomic gas, the heat capacity at constant volume \( C_v = \frac{3}{2} R \).

**Step 2: Calculate \( C_p \) and \( \gamma \)**
- The relationship between \( C_p \) and \( C_v \) is given by:
\[ C_p = C_v + R = \frac{3}{2} R + R = \frac{5}{2} R \]
- The ratio \( \gamma \) (gamma) is defined as:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{5}{2} R}{\frac{3}{2} R} = \frac{5}{3} \]

**Step 3: Use the Adiabatic Relation**
- For an adiabatic process, the relation between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
- Substituting the known values:
\[ 300 \cdot 8^{\frac{5}{3} - 1} = 250 \cdot V_2^{\frac{5}{3} - 1} \]
- Simplifying \( \frac{5}{3} - 1 = \frac{2}{3} \):
\[ 300 \cdot 8^{\frac{2}{3}} = 250 \cdot V_2^{\frac{2}{3}} \]

**Step 4: Calculate \( 8^{\frac{2}{3}} \)**
- Calculate \( 8^{\frac{2}{3}} \):
\[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4 \]

**Step 5: Substitute and Solve for \( V_2 \)**
- Substitute back into the equation:
\[ 300 \cdot 4 = 250 \cdot V_2^{\frac{2}{3}} \]
- This simplifies to:
\[ 1200 = 250 \cdot V_2^{\frac{2}{3}} \]
- Rearranging gives:
\[ V_2^{\frac{2}{3}} = \frac{1200}{250} = 4.8 \]

**Step 6: Solve for \( V_2 \)**
- To find \( V_2 \), raise both sides to the power of \( \frac{3}{2} \):
\[ V_2 = (4.8)^{\frac{3}{2}} = \sqrt{4.8^3} \]
- Calculate \( 4.8^3 \):
\[ 4.8^3 = 110.592 \quad \text{and} \quad \sqrt{110.592} \approx 10.5 \]

**Final Answer:**
- Therefore, the final volume \( V_2 \) is approximately \( 10.5 \, \text{L} \).