# 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.

Thank you very much for your useful answers.

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)? $$\frac{V_1}{V_2}=\frac{P_1\ T_2}{P_2\ T_1}?$$

Thank you in advance.

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? Mar 14, 2014 at 9:20
• If you pull the piston back out to the original volume do the pressure and temperature return to their original values? Mar 14, 2014 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. Mar 14, 2014 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. Mar 14, 2014 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. Mar 14, 2014 at 9:40

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