A gas expands \(0.25 \mathrm{~m}^{3}\) at Constant Pressure \(10^{3}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) the work done is (A) \(250 \mathrm{~J}\) (B) \(2.5\) erg (C) \(250 \mathrm{~W}\) (D) \(250 \mathrm{~N}\)

When a gas expands at constant pressure, the work done can be determined using the formula: \[W = P \Delta V\] where \(W\) is the work done, \(P\) is the constant pressure, and \(\Delta V\) is the change in volume.

We have been given the pressure \(P = 10^3 \frac{\mathrm N}{\mathrm m^2}\) and the change in volume \(\Delta V = 0.25 \mathrm m^3\). Substituting these into the formula, we get: \[W = (10^3 \frac{\mathrm N}{\mathrm m^2})(0.25 \mathrm m^3)\]

Now, we can calculate the work done. Multiplying the given values, we get: \[W = 250 \times 10^3 \mathrm{~J}\]

As we can see, our calculated value of work done is not equal to any of the given options. We can, however, convert the work to units given in some of the options: (A) 250 J is too small, given our calculated value. (B) \(1 \mathrm{~J} = 10^7 \mathrm{~erg}\), so \(250 \times 10^3 \mathrm{~J} = 250 \times 10^{10} \mathrm{~erg} = 2.5 \times 10^{12} \mathrm{~erg}\), not 2.5 erg. (C) A watt is a unit of power, not work, thus this option is not applicable. (D) 250 N is a unit of force, not work, thus this option is also not applicable. By analyzing the options, none of them match our calculated work done, leading us to believe the correct answer is not given . However, the closest option is (A) \(250 \mathrm{~J}\), if it were written as \(250 \times 10^3 \mathrm{~J}\).