Two moles of an ideal monoatomic gas occupies a volume V at 27$$^\circ$$C. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

We have an ideal mono-atomic gas with the following initial data:

Number of moles  $$n = 2$$

Initial volume  $$V_1 = V$$

Initial temperature  $$T_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$$

The gas undergoes an adiabatic expansion to a final volume

$$V_2 = 2V$$

For a reversible adiabatic process of an ideal gas we use the relation

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

where $$\gamma = \dfrac{C_P}{C_V}$$. For a mono-atomic ideal gas

$$C_V = \dfrac{3}{2}R,\quad C_P = \dfrac{5}{2}R\quad\Longrightarrow\quad\gamma = \dfrac{\tfrac{5}{2}R}{\tfrac{3}{2}R} = \dfrac{5}{3}$$

Hence,

$$\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$$

Applying the adiabatic relation to the initial and final states, we write

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

Substituting the known values,

$$300\,(V)^{\frac{2}{3}} = T_2\,(2V)^{\frac{2}{3}}$$

The common factor $$V^{\frac{2}{3}}$$ cancels, leaving

$$300 = T_2 \,(2)^{\frac{2}{3}}$$

So,

$$T_2 = \dfrac{300}{2^{\,\frac{2}{3}}}$$

We evaluate $$2^{\,\frac{2}{3}}$$ first. Taking the cube root of 2 we get approximately $$2^{1/3}\approx 1.26$$, and then squaring it:

$$2^{\,\frac{2}{3}} = (2^{1/3})^2 \approx (1.26)^2 \approx 1.59$$

Now,

$$T_2 \approx \dfrac{300}{1.59} \approx 189\text{ K}$$

This completes part (a).

To find the change in internal energy, we recall the formula for an ideal gas:

$$\Delta U = n C_V (T_2 - T_1)$$

For a mono-atomic gas $$C_V = \dfrac{3}{2}R$$, so

$$\Delta U = n\left(\dfrac{3}{2}R\right)(T_2 - T_1)$$

Substituting $$n = 2$$, $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$, $$T_2 = 189\text{ K}$$ and $$T_1 = 300\text{ K}$$, we get

$$\Delta U = 2 \times \dfrac{3}{2}R (189 - 300)$$

Simplifying the numeric coefficient first,

$$2 \times \dfrac{3}{2} = 3$$

So,

$$\Delta U = 3R (189 - 300) = 3R (-111)$$

Now substitute the value of $$R$$:

$$\Delta U = 3(8.314)\times(-111)\ \text{J}$$

$$\Delta U \approx 24.942 \times (-111)\ \text{J}$$

$$\Delta U \approx -2768.7\ \text{J}$$

Changing the unit to kilojoules,

$$\Delta U \approx -2.77\ \text{kJ}$$

Rounding suitably,

$$\Delta U \approx -2.7\ \text{kJ}$$

Hence, the gas cools to 189 K and its internal energy decreases by about 2.7 kJ.

Hence, the correct answer is Option D.