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 300 K, we can follow these steps:

### Step 1: Understand the relationship for adiabatic processes
For a reversible adiabatic process, the relationship between temperature and volume is given by:
\[ \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1} \]
where \( T_1 \) is the initial temperature, \( T_2 \) is the final temperature, \( V_1 \) is the initial volume, \( V_2 \) is the final volume, and \( \gamma \) is the heat capacity ratio (Cp/Cv).

### Step 2: Identify the known values
From the problem, we have:
- \( V_1 = 16 \, L \)
- \( V_2 = 2 \, L \)
- \( T_1 = 300 \, K \)

### Step 3: Calculate the value of \( \gamma \)
For a monoatomic ideal gas, the heat capacities are:
- \( C_v = \frac{3R}{2} \)
- \( C_p = C_v + R = \frac{3R}{2} + R = \frac{5R}{2} \)

Thus, the ratio \( \gamma \) is:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} \]

### Step 4: Substitute the values into the adiabatic equation
Now we can substitute the known values into the adiabatic equation:
\[ \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1} \]

Substituting the values:
\[ \frac{300}{T_2} = \left(\frac{2}{16}\right)^{\frac{5}{3} - 1} \]

### Step 5: Simplify the equation
First, simplify \( \frac{2}{16} \):
\[ \frac{2}{16} = \frac{1}{8} \]

Now calculate \( \gamma - 1 \):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]

Now we can rewrite the equation:
\[ \frac{300}{T_2} = \left(\frac{1}{8}\right)^{\frac{2}{3}} \]

### Step 6: Calculate \( \left(\frac{1}{8}\right)^{\frac{2}{3}} \)
Calculating \( \left(\frac{1}{8}\right)^{\frac{2}{3}} \):
\[ \left(\frac{1}{8}\right)^{\frac{2}{3}} = \frac{1^{2/3}}{8^{2/3}} = \frac{1}{4} \]

### Step 7: Solve for \( T_2 \)
Now substituting back:
\[ \frac{300}{T_2} = \frac{1}{4} \]

Cross-multiplying gives:
\[ 300 = \frac{T_2}{4} \]

Thus,
\[ T_2 = 300 \times 4 = 1200 \, K \]

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