Recently, I came across a problem which involved the above setup. A gas is filled in the cylinder shown in the figure. The two friction-less pistons(red) are joined by an in-extensible string(blue). Answer the following:

1. If this system is placed under isolated vacuum conditions(space), the pistons will?
2. If the gas is heated, the pistons will?

Comment below with what you think will happen!

Note by Raghav Vaidyanathan
6 years, 3 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

since the pressure will be same............but the area of right piston is greater ..the pressure force on that will be greater.......................so it will move rightwards

- 6 years, 3 months ago

But all this is internal force! If the piston moves rightwards, the center of mass will also move towards the right, which is is impossible for an internal force to do!

- 6 years, 3 months ago

the piston shall move.....as a pressure force would be exerted on the vertical wall of the container also ...on the left part ..the vertical wall which is present to make connection between the cylinder of the smaller piston and the larger piston............... the force on this wall would provide a reaction....which would be an external force............

- 6 years, 3 months ago

Good job! That is the answer that I was expecting. It took me a while for me to realize that the gas exerts a force on the vertical walls also.

- 6 years, 3 months ago

yes the situation was really tricky...........

- 6 years, 3 months ago

- 6 years, 3 months ago

Yash gave the right answer. As a stepping-stone in analysis, consider the case where we only have the right piston, enclosing a certain quantity of gas inside a cylinder with a bottom. What happens when the apparatus is exposed to high vacuum? The piston will move. In this case, the smaller piston also wants to move out, but the force pushing the smaller piston is overcome by the force pushing the larger.

- 6 years, 3 months ago

Thank you! My concern was that momentum wasn't getting conserved. But I later found out that it is. In the single piston case that you mentioned, the piston will move outward and the container will move backward, thus conserving momentum. Something similar happens here too, the connected pistons move toward the right and the container moves to the left.

- 6 years, 3 months ago

I have understood this problem, it is an interesting one to analyze.

- 6 years, 3 months ago

Pistons will surely move right. But still the momentum will remain conserved. There is no net motion of the center of mass of system as a whole . Thats all I saw in the problem above

- 6 years, 3 months ago