A A mole of helium gas initially at stp undergoes an isovolumetric process in which its pressure falls to half its initial value. The work done by gas is

To determine the work done by a mole of helium gas during an isovolumetric process where its pressure decreases to half its initial value, we need to understand a few key concepts about gas behavior and thermodynamics.

Understanding Isovolumetric Processes

An isovolumetric process, also known as an isochoric process, occurs at constant volume. This means that the gas does not expand or compress, and therefore, no work is done by the gas. In thermodynamics, work (W) done by a gas is defined by the equation:

• W = PΔV

Where:

• P is the pressure of the gas.
• ΔV is the change in volume.

Since the volume remains constant in an isovolumetric process, the change in volume (ΔV) is zero. Thus, we can conclude:

• W = P * 0 = 0

Analyzing the Situation

In your scenario, the helium gas starts at standard temperature and pressure (STP), which is defined as 1 atm (approximately 101.3 kPa) and 0 degrees Celsius (273.15 K). When the pressure of the gas falls to half its initial value, it drops to 0.5 atm (about 50.65 kPa). However, since the volume does not change during this process, the work done by the gas remains zero.

Why No Work is Done

To visualize this, think of a balloon filled with helium. If you were to keep the balloon at a fixed size (isovolumetric) and simply let some of the gas escape, the pressure inside the balloon would decrease, but the balloon itself wouldn’t change shape or size. Since the volume is constant, there’s no expansion against an external pressure, and thus, no work is performed.

Final Thoughts

In summary, regardless of the changes in pressure or temperature during an isovolumetric process, if the volume remains constant, the work done by the gas is always zero. This principle is fundamental in thermodynamics and helps us understand how gases behave under different conditions.