1. Given data

• Initial temperature: $T_1 = 300\,\text{K}$
• Number of moles: $n=1$
• Diatomic ideal gas with $\gamma = 1.4 = \dfrac{7}{5}$
• First process: adiabatic compression from $V_1$ to $V_2 = \frac{V_1}{16}$
• Second process: isobaric expansion from $V_2$ to $V_3 = 2V_2$

We need the final temperature $T_3$.

2. Temperature after adiabatic compression

For a quasi-static adiabatic process of an ideal gas, $TV^{\gamma-1} = \text{constant}$

So, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$

Hence, $T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$

Now, $\frac{V_1}{V_2} = 16, \qquad \gamma -1 = 1.4-1 = 0.4 = \frac{2}{5}$

Therefore, $T_2 = 300\times 16^{0.4}$

Since $16 = 2^4 \implies 16^{0.4} = (2^4)^{2/5} = 2^{8/5}$

Numerically, $16^{0.4} \approx 3.0314$

Thus, $T_2 \approx 300\times 3.0314 = 909.4\,\text{K}$

3. Temperature after isobaric expansion

For an isobaric process, $\frac{T}{V} = \text{constant}$

So, $\frac{T_3}{T_2} = \frac{V_3}{V_2}$

Given $V_3 = 2V_2$, therefore $T_3 = 2T_2$

Hence, $T_3 \approx 2\times 909.4 = 1818.8\,\text{K}$

4. Final answer

To the nearest integer, $T_3 \approx 1819\,\text{K}$

5. Comparison with stored answer

Stored correct answer: $1818 \text{ to } 1819$

Our derived answer is $1819$, which lies in the given accepted range. So it agrees with the stored answer.