0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m$$^3$$ at 1000 K. Given, R is the gas constant in JK$$^{-1}$$mol$$^{-1}$$, x is:

We start with the ideal-gas equation, which states that for any ideal gas the relation $$PV = nRT$$ holds, where $$P$$ is the pressure, $$V$$ is the volume, $$n$$ is the total number of moles present, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

In the vessel we have two gases together, so the total number of moles is the sum of the individual moles: $$n = 0.5 + x$$.

The numerical data given are: $$P = 200\ \text{Pa}, \quad V = 10\ \text{m}^3, \quad T = 1000\ \text{K}$$, while $$R$$ is to be kept as a symbol. Substituting all these directly into $$PV = nRT$$ gives

$$200 \times 10 = (0.5 + x)\, R \times 1000.$$

Simplifying the left-hand side first, we obtain

$$200 \times 10 = 2000,$$

so the equation becomes

$$2000 = (0.5 + x)\, 1000 R.$$

To isolate the bracket, we divide both sides by $$1000R$$:

$$\frac{2000}{1000R} = 0.5 + x.$$

The fraction on the left simplifies because $$2000/1000 = 2$$, hence

$$\frac{2}{R} = 0.5 + x.$$

Now we solve for $$x$$ by subtracting $$0.5$$ from both sides:

$$x = \frac{2}{R} - 0.5.$$

Writing the decimal $$0.5$$ as the fraction $$\dfrac{1}{2}$$ and bringing the two fractions to a common denominator $$2R$$, we get

$$x = \frac{2}{R} - \frac{1}{2} = \frac{4}{2R} - \frac{R}{2R} = \frac{4 - R}{2R}.$$

This expression matches Option C in the list. Hence, the correct answer is Option C.