Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a problem with an isochor transformation. Me and my group of study made an experiment that want to check Gay-Lussac’s law. We registered the equilibrium states and fitted the $P = nRT / V$, in the $P(T)$ graph. The moles found are $n_f=(5.63 \times 10^{-3} \pm 1.24 \times 10^{-4})$ mol. The problem is that, initial moles are $n_i=(3.19 \times 10^{-3} \pm 6.97 \times 10^{-5})$ mol. How can the initial moles be smaller?

Details of the experiment: We started stated with a certain number of moles of air inside a metal cylinder closed with a rubber top and connected, through it, to a temperature and pressure sensor. The starting gas conditions are $V, P_0, T_0$ assuming the volume to be constant (the container is, in a good approximation, hermetic).

We heated an amount of water in a calorimeter with a resistance and a temperature $T' \gg T_0$, then we put the cylinder in contact with the hot water plotting the trend of $\Delta P-T$ on a graph. From a certain point the trend gets linear, and with a fit we calculated the slope $m = nR/V$ and from the latter we calculated the number of moles $n$ assuming the air as an ideal gas because of the higher temperature.

We checked the consistency between this value and the one calculated from the initial conditions $n' = P_0V/RT_0$ and surprisingly this value is smaller than the one calculated from the fit. Now what is really happening? the difference $n'-n < 0$ suggests that the number of moles in the cylinder is growing? Or maybe it might be that one of our assumptions is incorrect.

share|improve this question

1 Answer 1

I would independantly check your first result e.g. measure the volume by filling your container with water and measuring the weight change. One mole of an ideal gas is 0.0224m$^3$, so that gives you a third way to calculate the number of moles present.

You say:

From a certain point the trend gets linear

Air is pretty close to an ideal gas so a PT graph should be linear. If you're getting a curve that would worry me. Maybe you're heating too fast and the air inside the container isn't all at the same temperature.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.