An ideal gas at initial temperature 300 K is compressed adiabatically (`gamma = 1.4`) to `(1/16)^(th)` of its initial volume. The gas is then expanded isobarically to double its volume. Then final temperature of gas round to nearest integer is:

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the Initial Conditions
- The initial temperature \( T_1 = 300 \, K \).
- The gas is compressed adiabatically to \( \frac{1}{16} \) of its initial volume.

### Step 2: Use the Adiabatic Relation
For an adiabatic process, we can use the relation:
\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
Where:
- \( \gamma = 1.4 \)
- \( V_2 = \frac{1}{16} V_1 \)

Thus, we can express \( V_1/V_2 \) as:
\[ \frac{V_1}{V_2} = 16 \]

### Step 3: Substitute Values into the Adiabatic Relation
Now, substituting the values into the adiabatic relation:
\[ \frac{T_2}{300} = 16^{(1.4 - 1)} = 16^{0.4} \]

### Step 4: Calculate \( 16^{0.4} \)
Calculating \( 16^{0.4} \):
\[ 16^{0.4} = (2^4)^{0.4} = 2^{4 \times 0.4} = 2^{1.6} \]
Using a calculator or logarithmic tables, we find:
\[ 2^{1.6} \approx 3.17 \]

### Step 5: Calculate \( T_2 \)
Now, substituting back to find \( T_2 \):
\[ T_2 = 300 \times 3.17 \approx 951 \, K \]

### Step 6: Use the Isobaric Process Relation
Next, the gas is expanded isobarically (constant pressure) to double its volume:
\[ \frac{V_3}{V_2} = 2 \]
Since volume is proportional to temperature in an isobaric process:
\[ \frac{T_3}{T_2} = \frac{V_3}{V_2} = 2 \]

### Step 7: Calculate \( T_3 \)
Now, substituting \( T_2 \) into the equation:
\[ T_3 = 2 \times T_2 = 2 \times 951 \approx 1902 \, K \]

### Step 8: Round to Nearest Integer
Finally, rounding \( T_3 \) to the nearest integer gives:
\[ T_3 \approx 1902 \, K \]

### Final Answer
The final temperature of the gas, rounded to the nearest integer, is:
\[ \boxed{1902} \]