Question

Refer to an imaginary "frictionless" piston. This piston is capable of moving up or down within the container's walls without any resistance (assume there is no friction between the walls of the container and the piston). In addition, there is a complete seal around the piston, which prevents any particles from getting into or out of the container. In figure (a), the piston is at a resting position, with the pressure inside equal to the pressure outside. The temperatures inside and outside are identical. In figure (b), the piston is drawn out so that the volume inside the container is twice as great as in figure (a). Describe what would happen to the pressure inside the containerand explain why this would occur.

Answer 1

ANSWER and EXPLANATION

To explain this, we have to apply Boyle's law which states that:

This implies that as pressure increases, volume decreases, and as volume increases pressure decreases.

It is given mathematically as:

`\begin{gathered} P\propto(1)/(V) \\ \\ P=(k)/(V) \\ \\ P_1V_1=P_2V_2 \end{gathered}`

where k = constant of proportionality

P = pressure

V = volume

In the given figures, the piston is drawn out so that the volume inside the container is doubled.

From the equation, if V2 is twice as much as V1, it implies that:

`P_1V_1=P_2(2V_1)`

Solve for P2 in the equation:

`\begin{gathered} P_2=(P_1V_1)/(2V_1) \\ \\ P_2=(P_1)/(2) \end{gathered}`

Therefore, as we see from the solution above, the pressure inside the container will be half of what it originally was.

Hence, if the volume inside the container is twice as great as before, the pressure is halved.