A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $$\gamma = 1.40$$, the work done by air is: ($$3^{1.4} = 4.6555$$) [Take air to be an ideal gas]

We start by writing all data in convenient SI units. The initial volume of air is $$V_1 = 1\;\text{L} = 1 \times 10^{-3}\;\text{m}^3$$ and the final volume is $$V_2 = 3\;\text{L} = 3 \times 10^{-3}\;\text{m}^3$$. At STP the initial pressure is $$P_1 = 1\;\text{atm} = 1.013 \times 10^{5}\;\text{Pa}$$ and the initial temperature is $$T_1 = 273\;\text{K}$$. For air the adiabatic index is $$\gamma = 1.40$$.

An adiabatic process for an ideal gas obeys the relation $$P V^{\gamma} = \text{constant}.$$ Using this, the final pressure $$P_2$$ can be obtained from

$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\,,$$

so

$$P_2 = P_1 \left(\dfrac{V_1}{V_2}\right)^{\gamma}.$$

Here $$\dfrac{V_1}{V_2} = \dfrac{1}{3}$$, hence

$$P_2 = 1.013 \times 10^{5}\;\text{Pa}\;\times \left(\dfrac{1}{3}\right)^{1.40}.$$

The question supplies $$3^{1.4} = 4.6555$$, therefore

$$\left(\dfrac{1}{3}\right)^{1.4} = \dfrac{1}{3^{1.4}} = \dfrac{1}{4.6555} = 0.2148.$$

Substituting, we get

$$P_2 = 1.013 \times 10^{5}\;\text{Pa} \times 0.2148 = 2.1768 \times 10^{4}\;\text{Pa}.$$

Now we recall the formula for the work done during an adiabatic process for an ideal gas:

$$W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}.$$

First calculate the two pressure-volume products (remembering that $$1\;\text{Pa}\cdot\text{m}^3 = 1\;\text{J}$$):

$$P_1 V_1 = \bigl(1.013 \times 10^{5}\;\text{Pa}\bigr)\bigl(1 \times 10^{-3}\;\text{m}^3\bigr) = 101.3\;\text{J},$$

$$P_2 V_2 = \bigl(2.1768 \times 10^{4}\;\text{Pa}\bigr)\bigl(3 \times 10^{-3}\;\text{m}^3\bigr) = 65.304\;\text{J}.$$

The difference of these two terms is

$$P_1 V_1 - P_2 V_2 = 101.3\;\text{J} - 65.304\;\text{J} = 35.996\;\text{J}.$$

Now divide by $$\gamma - 1$$. Since $$\gamma = 1.40$$, we have $$\gamma - 1 = 0.40$$, so

$$W = \dfrac{35.996\;\text{J}}{0.40} = 89.99\;\text{J} \approx 90.5\;\text{J}.$$

Hence, the correct answer is Option B.