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, we will follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process for an ideal gas, the relationship between the initial and final states can be described by the equation:

\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

where:
- \( T_1 \) = initial temperature
- \( V_1 \) = initial volume
- \( T_2 \) = final temperature after adiabatic compression
- \( V_2 \) = final volume after adiabatic compression
- \( \gamma \) = heat capacity ratio (given as 1.4)

### Step 2: Apply the Adiabatic Condition
Given:
- \( T_1 = 300 \, K \)
- \( V_2 = \frac{1}{16} V_1 \)

Using the adiabatic condition:

\[ 300 \cdot V_1^{\gamma - 1} = T_2 \cdot \left(\frac{1}{16} V_1\right)^{\gamma - 1} \]

Substituting \( \gamma = 1.4 \):

\[ 300 \cdot V_1^{0.4} = T_2 \cdot \left(\frac{1}{16}\right)^{0.4} V_1^{0.4} \]

### Step 3: Simplify the Equation
Cancel \( V_1^{0.4} \) from both sides:

\[ 300 = T_2 \cdot \left(\frac{1}{16}\right)^{0.4} \]

Now calculate \( \left(\frac{1}{16}\right)^{0.4} \):

\[ \left(\frac{1}{16}\right)^{0.4} = \left(16^{-1}\right)^{0.4} = 16^{-0.4} = 2^{-1.6} = \frac{1}{2^{1.6}} \approx \frac{1}{3.17} \approx 0.316 \]

### Step 4: Solve for \( T_2 \)
Now substitute back into the equation:

\[ 300 = T_2 \cdot 0.316 \]

Solving for \( T_2 \):

\[ T_2 = \frac{300}{0.316} \approx 948.10 \, K \]

### Step 5: Isobaric Expansion
The gas is then expanded isobarically to double its volume. The relationship for an ideal gas during an isobaric process is:

\[ \frac{T_2}{T_1} = \frac{V_2}{V_1} \]

Here, since the volume doubles, we have:

\[ T_3 = T_2 \cdot 2 \]

Substituting \( T_2 \):

\[ T_3 = 948.10 \cdot 2 \approx 1896.20 \, K \]

### Step 6: Round to Nearest Integer
Finally, rounding \( T_3 \) to the nearest integer gives:

\[ T_3 \approx 1896 \, K \]

### Final Answer
The final temperature of the gas, rounded to the nearest integer, is **1896 K**.

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