A cylindrical container of volume $$4.0 \times 10^{-3}$$ m$$^3$$ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is:
[Take gas constant as 8.3 J mol$$^{-1}$$ K$$^{-1}$$]

For an ideal gas, the equation of state is stated as $$pV = nRT,$$ where $$p$$ is the pressure, $$V$$ is the volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

We have one mole of hydrogen and two moles of carbon dioxide. So the total number of moles present in the cylinder is

$$n = 1 + 2 = 3 \text{ mol}.$$

The gas constant is given as $$R = 8.3 \text{ J mol}^{-1}\text{ K}^{-1}.$$

The absolute temperature of the mixture is given to be $$T = 400 \text{ K}.$$

The volume of the cylindrical container is

$$V = 4.0 \times 10^{-3} \text{ m}^3.$$

Now we substitute these values into the ideal-gas equation. Rearranging the formula for pressure, we get

$$p = \frac{nRT}{V}.$$

Substituting $$n = 3 \text{ mol},\; R = 8.3 \text{ J mol}^{-1}\text{ K}^{-1},\; T = 400 \text{ K},$$ and $$V = 4.0 \times 10^{-3} \text{ m}^3,$$ we obtain

$$p = \frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}.$$

First, multiply the numerator step by step:

$$8.3 \times 400 = 3320,$$

so

$$3 \times 3320 = 9960.$$

Hence the numerator equals $$9960.$$

Next we divide by the volume:

$$p = \frac{9960}{4.0 \times 10^{-3}}.$$

Recognise that dividing by $$4.0 \times 10^{-3}$$ is the same as multiplying by $$\frac{1}{4.0 \times 10^{-3}} = \frac{1}{0.004} = 250.$$ Therefore,

$$p = 9960 \times 250.$$

Carrying out this multiplication:

$$9960 \times 250 = 2\,490\,000.$$

Expressing this in scientific notation,

$$2\,490\,000 \text{ Pa} = 2.49 \times 10^{6} \text{ Pa}.$$

Writing the same value with one decimal place different,

$$2.49 \times 10^{6} \text{ Pa} = 24.9 \times 10^{5} \text{ Pa}.$$

This matches option C of the given choices.

Hence, the correct answer is Option 3.