Starting at temperature $$300K$$, one mole of an ideal diatomic gas $$(\gamma = 1.4)$$ is first compressed adiabatically from volume $$V_1$$ to $$V_2 = \frac{V_1}{16}$$. It is then allowed to expand isobarically to volume $$2V_2$$. If all the processes are the quasi-static then the final temperature of the gas (in $$°K$$) is (to the nearest integer) ___________.

We are told that one mole of an ideal di-atomic gas is initially at temperature $$T_1 = 300\;{\rm K}$$ and volume $$V_1$$. Its ratio of specific heats is given as $$\gamma = 1.4$$. The gas is first compressed adiabatically to a smaller volume $$V_2$$ where $$V_2 = \dfrac{V_1}{16}$$.

For any reversible (quasi-static) adiabatic process of an ideal gas, we recall the relation

$$T\,V^{\gamma-1} = \text{constant}.$$

This means that for the initial state (1) and the state just after the adiabatic compression (2) we must have

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

Solving for $$T_2$$ gives

$$T_2 \;=\; T_1 \left(\dfrac{V_1}{V_2}\right)^{\gamma-1}.$$

We know $$\dfrac{V_1}{V_2} = 16$$ and $$\gamma - 1 = 1.4 - 1 = 0.4$$, so

$$T_2 \;=\; 300\;{\rm K}\;\times\; 16^{0.4}.$$

We next evaluate $$16^{0.4}$$. Noting that $$16 = 2^4$$, we have

$$16^{0.4} = (2^4)^{0.4} = 2^{4 \times 0.4} = 2^{1.6}.$$

Using $$2^{1.6} \approx 3.03$$, we obtain

$$T_2 \;\approx\; 300 \times 3.03 \;=\; 909\;{\rm K}.$$

The second step in the problem is an isobaric (constant-pressure) expansion from volume $$V_2$$ to volume $$V_3 = 2V_2$$. For a constant-pressure process of an ideal gas we invoke the ideal-gas law in the form

$$\dfrac{T}{V} = \text{constant (when $P$ is constant)},$$

so that

$$\dfrac{T_2}{V_2} = \dfrac{T_3}{V_3}.$$

Solving for the final temperature $$T_3$$ we get

$$T_3 \;=\; T_2 \left(\dfrac{V_3}{V_2}\right).$$

Since $$V_3 = 2V_2$$, this reduces to

$$T_3 \;=\; T_2 \times 2.$$

Substituting the value of $$T_2$$ found above, we find

$$T_3 \;\approx\; 2 \times 909\;{\rm K} \;=\; 1818\;{\rm K}.$$

Rounding to the nearest integer, the final temperature becomes $$1819\;{\rm K}$$.

Hence, the correct answer is Option 1819.