# Doubts about the ideal gas law

EDITED AND SOLUTION: In fact my stupid mistake was to take the wrong value for my $P_1$ and I was getting an 1.2998 that finally is a 1,9178 within just a difference about 7%. So at the end they are not equal but it is not a $1.9\neq1.3$, but $1,9\neq1,9278$.

In spite of everything, it has been profitable to me to discuss the possible conditions where the ideal gas law can be a good aproximation.

I have made an experiment that consist of a gas inside a cylinder and I have a piston. I have this experiment connected to a computer and I can graph the temperature and the pressure. This system can lose temperature so when I push in the cylinder I can see that my temperature increases and estabilizes to my lab temperature after 5 seconds.

So, my question is:

Why am I getting a factor $\frac{V_1}{V_2}=1,9 \pm 0,1$ and a factor $\frac{P_1\ T_2}{P_2\ T_1}=1,2998\pm0,0001$?

I know that they have to be different but I don't know why, so why I shouldn't be expecting them to be equal (taking in account the uncertainties)? $$\frac{V_1}{V_2}=\frac{P_1\ T_2}{P_2\ T_1}?$$

PLUS! I have added the image of the experiment. I pushed in the piston and stay making force for about 20 sec until I released the piston free M.

• Are You sure that Your cylinder is gas-tight and the gas is not espcaping? – Wojciech Mar 14 '14 at 9:20
• If you pull the piston back out to the original volume do the pressure and temperature return to their original values? – John Rennie Mar 14 '14 at 9:31
• Wojciech: Yes! We are using a PASCO system that is not supposed to have any sort of escapes. In fact I think that it's more like a theoretical thermodynamic problem that than a bad praxis. – Marc C Mar 14 '14 at 9:32
• John Rennie: the temperature returns to the initial value (lab temperature) without any move. If I pull the piston back out to the original volume the temperature decreases for 5 seconds until it stabilizes to the lab temperature. The pressure returns to the original value. – Marc C Mar 14 '14 at 9:36
• The gas You are using is a real gas, not ideal gas, so that ideal gas law may not be accurate especially if Your gas is not monoatomic at high temperature and low pressure. – Wojciech Mar 14 '14 at 9:40

$pv=RT$ gas equation
$T_1=T_2$ so $p_1v_1=p_2v_2$ or $v_1/v_2=p_2/p_1$