One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27$$^\circ$$C. The work done on the gas will be:

We have one mole of an ideal mono-atomic gas, so $$n = 1$$. The process is carried out isothermally at room temperature, which is given as $$27^{\circ}\text{C}$$. Converting to kelvin, $$T = 27 + 273 = 300\,\text{K}$$.

For an isothermal process in an ideal gas, the work done by the gas is found from the well-known formula

$$W_{\text{by}} = nRT \ln\!\left(\frac{V_f}{V_i}\right).$$

Because the temperature is constant, the ideal-gas equation $$PV = nRT$$ tells us that $$P \propto \dfrac1V$$. Hence

$$\frac{V_f}{V_i} = \frac{P_i}{P_f}.$$

The statement “pressure is doubled” means $$P_f = 2P_i$$. Substituting this ratio we get

$$\frac{V_f}{V_i} = \frac{P_i}{2P_i} = \frac12.$$

Putting this value into the work expression gives

$$W_{\text{by}} = nRT \ln\!\left(\frac12\right).$$

Now, $$\ln\!\left(\dfrac12\right) = -\ln 2$$, so

$$W_{\text{by}} = nRT (-\ln 2) = -\,nRT \ln 2.$$

But we are asked for the work done on the gas, which is the negative of the work done by the gas:

$$W_{\text{on}} = -W_{\text{by}} = nRT \ln 2.$$

With $$n = 1$$ and $$T = 300\,\text{K}$$, we obtain

$$W_{\text{on}} = (1)\,R\,(300)\,\ln 2 = 300R \ln 2.$$

Hence, the correct answer is Option D.