Question

One mole of helium gas, initially at STP (p₁ = 1 atm, T₁ = 0°C), undergoes an isovolumetric process in which its pressure falls to half its initial value. The work done by the gas is:

• A. 101 J
• B. 51 J
• C. 23 J
• D. 0 J

Hint:

In an isovolumetric process, the volume remains constant, meaning no work is done by the gas.

Answer 1

The Correct Option is D

Step 1: Recall how work is defined for a gas.
The work done by a gas as it changes volume is the integral of pressure over the volume change:
\[ W = \int_{V_1}^{V_2} P \, dV \]

Step 2: Apply the isovolumetric condition.
Isovolumetric means the volume stays fixed throughout the process, so \( V_1 = V_2 \). The limits of the integral become equal to each other.

Step 3: Evaluate the integral.
An integral taken from a point back to the same point is always zero, regardless of what the pressure is doing in between:
\[ W = \int_{V_1}^{V_1} P \, dV = 0 \]
This holds here even though the pressure itself drops to half its starting value, since it is the volume change, not the pressure change, that determines work.

Step 4: Final Answer.
\[ \boxed{W = 0\,J} \]

Answer 2

A helpful way to picture this is on a pressure-volume diagram, where the work done by a gas is represented by the area under the curve it traces as it moves from one state to another.

1. 101 J: On a pressure-volume diagram, an isovolumetric process is drawn as a straight vertical line, since the volume coordinate never changes even as the pressure drops. A vertical line encloses no horizontal width, so it cannot represent a nonzero work value like this.
2. 51 J: Same reasoning applies here. Any nonzero work would need the process to trace out some horizontal distance on the diagram, corresponding to an actual volume change, which a vertical line does not have.
3. 23 J: Still assumes some sideways movement on the diagram. A purely vertical line has zero width under it, so this value cannot be correct either.
4. 0 J: A vertical line on a pressure-volume diagram has zero horizontal extent, and area under a line with zero width is zero. Since work equals that area, dropping the pressure at constant volume gives exactly zero work.

Whether pictured as an integral or as an area on a pressure-volume graph, the story is the same: no change in volume means no area, and no area means no work.

The correct answer is 0 J.