Calculate the final temperature of a monoatomic ideal gas that is compressed reversible and adiabatically from `16L` to `2L` at `300 K :`

Options

A. 600 K
B. 1044.6 K
C. 1200 K
D. 2400 K

Explanation

To calculate the final temperature of a monoatomic ideal gas that is compressed reversibly and adiabatically from 16 L to 2 L at an initial temperature of 300 K, we can follow these steps:

### Step 1: Identify the Given Values
- Initial Volume (V1) = 16 L
- Final Volume (V2) = 2 L
- Initial Temperature (T1) = 300 K

### Step 2: Determine the Value of Gamma (γ)
For a monoatomic ideal gas, the ratio of specific heats (γ) is given by:
\[ \gamma = \frac{C_p}{C_v} \]
For monoatomic gases:
- \(C_v = \frac{3R}{2}\)
- \(C_p = C_v + R = \frac{3R}{2} + R = \frac{5R}{2}\)

Thus,
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} \]

### Step 3: Use the Adiabatic Condition
For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
We can rearrange this to find \(T_2\):
\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]

### Step 4: Substitute the Known Values
Now substitute the known values into the equation:
\[ T_2 = 300 \left(\frac{16}{2}\right)^{\frac{5}{3} - 1} \]
Calculating the fraction:
\[ \frac{16}{2} = 8 \]
Now calculate \(\gamma - 1\):
\[ \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]

### Step 5: Calculate \(T_2\)
Now plug this back into the equation:
\[ T_2 = 300 \cdot 8^{\frac{2}{3}} \]
Calculating \(8^{\frac{2}{3}}\):
\[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4 \]
Thus,
\[ T_2 = 300 \cdot 4 = 1200 \text{ K} \]

### Final Answer
The final temperature \(T_2\) of the gas after the adiabatic compression is:
\[ \boxed{1200 \text{ K}} \]