Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is:

We are told that five moles of an ideal gas, initially at a pressure of 1 bar and a temperature of 298 K, is suddenly allowed to expand into a vacuum until its volume becomes double of the initial volume. Because the surroundings on the other side of the partition is a vacuum, the external pressure that opposes the expansion is zero.

In thermodynamics, the elemental work done by the system during a small change in volume is given by the formula

$$\mathrm dW = P_{\text{ext}}\,\mathrm dV$$

where $$P_{\text{ext}}$$ is the external (opposing) pressure. To obtain the total work $$W$$, we integrate the above expression from the initial volume $$V_1$$ to the final volume $$V_2$$:

$$W = \int_{V_1}^{V_2} P_{\text{ext}}\,\mathrm dV$$

For a free expansion into a vacuum, we have

$$P_{\text{ext}} = 0 \quad\text{for all values of } V$$

Substituting $$P_{\text{ext}} = 0$$ into the integral, we get

$$W = \int_{V_1}^{V_2} 0 \,\mathrm dV$$

Since the integrand is zero everywhere, the entire integral evaluates to

$$W = 0$$

Thus, even though the volume doubles and the process involves five moles of gas, the work done on the surroundings is zero because there is no opposing external pressure.

Examining the given options, only Option D states a work of zero.

Hence, the correct answer is Option D.