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

I pushed in the piston and stay making force for about 20 sec until I released the piston free


  • $\begingroup$ Are You sure that Your cylinder is gas-tight and the gas is not espcaping? $\endgroup$
    – Wojciech
    Mar 14, 2014 at 9:20
  • $\begingroup$ If you pull the piston back out to the original volume do the pressure and temperature return to their original values? $\endgroup$ Mar 14, 2014 at 9:31
  • $\begingroup$ 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. $\endgroup$ Mar 14, 2014 at 9:32
  • $\begingroup$ 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. $\endgroup$ Mar 14, 2014 at 9:36
  • 2
    $\begingroup$ 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. $\endgroup$
    – Wojciech
    Mar 14, 2014 at 9:40

1 Answer 1


The pressures are moderate so ideal gas is a good assumption. Moreover the temperatures equalize quickly so you can treat your problem isothermally with regard to the end states:

$pv=RT$ gas equation

$T_1=T_2$ so $p_1v_1=p_2v_2$ or $v_1/v_2=p_2/p_1$

If I take your pressure reading between 0 and 5s, I estimate it to be 92kPa. For the compressed state about 163kPa. So the ratio would be around 1.77. That's not too far away from your volume ratio of 1.9 and might easily be explained by calculating the errors. The reason for the discrepancy between start and end pressure seems to be due to the resistance of the piston, which if you let it go, does not travel back to the same position as you started off.

  • $\begingroup$ That was trully decisive. $\endgroup$ Mar 15, 2014 at 10:52

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