# How constant volume gas thermomether keeps the volume constant?

So the level of mercury on the left is being kept down by the pressure of the gas, correct? When the temperature rises, shouldn't the mercury level on the left go down? And when the mercury level go down, the volume filled with gas increase, so where is the constante volume that gives name to the device? Help me please, I need to understand this.

The answer is that when the temperature rises of the gas, the pressure goes up and it pushes the mercury down, but then the tube on the right is lifted up by the person operating the experiment to increase the pressure on the gas and return it to its original volume - so the height h increases.

The reason that there is flexible tube at the bottom fo the diagram is so that the open containter on the right hand side can be raised or lowered.

As $h$ is varied the pressure varies and the volume of the gas remains constant.

The pressure unit Torr is equivalent to mmHg - or millimeter height of mercury, which is why the $h$ is so important on the right hand side of the diagram as the pressure in the gas on the left is equal to atmospheric pressure plus the pressure from the mercury - make $h$ negative and you can reduce the pressure in the measuring chamber for lower temperatures.

• Oh now I understand. I wasn't sure about the function of the flexible tube, now makes sense. They should add this on textbooks, most of them don't mention that a person need to be involved in the experiment. Thank you. – João Pedro Dec 5 '14 at 16:39
• @JoãoPedro - if you think that is your question answered then you might want to indicate that the question is answered so that other people on the site see when they look at your question - see physics.stackexchange.com/help/someone-answers - glad that the answer was useful. – tom Dec 5 '14 at 16:41
• I'm new to "stacks" didn't know you could indicate that, lol. Thought I could only thumbs down or up and only when I got enough rep! – João Pedro Dec 5 '14 at 16:46
• @JoãoPedro - no problem - and thanks for marking the question answered - I suggest that you look at your questions on the Maths SE and see if you want to mark any of them as answered as well. Welcome to SE!! – tom Dec 5 '14 at 16:48