Final answer:
No work is done in a thermodynamic process if there is constant volume. This is because work is calculated as pressure multiplied by the change in volume, and with constant volume there is no change, hence no work. Constant pressure does not guarantee no work is done, as the volume could still change.
Step-by-step explanation:
In the context of physical sciences, work is done when a force causes a displacement of an object. Work can be calculated by the equation W = F × d × cos(θ), where W is the work done, F is the force applied, d is the displacement, and θ is the angle between the force and displacement vectors. For the purpose of this question, the conditions under which no work is done can be interpreted through the lens of thermodynamics, specifically during processes involving gases.
With reference to an ideal gas in a thermodynamic process, no work is done when there is constant volume. Constant volume implies that there is no change in volume (ΔV = 0), and since the definition of work in thermodynamics is W = p×ΔV (where p is pressure and ΔV is the change in volume), the work done would be zero if ΔV is zero. Hence, constant volume meets the condition for no work being done during a thermodynamic process.
In contrast, constant pressure by itself does not assure that no work is done because the volume could still change under constant pressure, as might happen in an isobaric process. During an isobaric process, where pressure is held constant while the gas volume changes, work is indeed done. Similarly, work can also occur under constant temperature conditions, as explained by Charles's Law where volume is directly proportional to temperature when the pressure is constant. Therefore, 'both A and C' does not meet the condition for no work as constant pressure doesn't inherently mean there's no change in volume.