1
$\begingroup$

Let's say that I have two boxes with both volumes equal $V$, filled with an air.

Air in first box have pressure $p_1$ and temperature $T_1$, likewise air in second box have pressure $p_2$ and temperature $T_2$.

Suppose this is ideal gas, then from Clapeyron equation we have:

$$ p_1 V = n_1 R T_1 $$ $$ p_2 V = n_2 R T_2 $$

Thus, after simple transformation:

$$ \frac{p_1 V} {R T_1} = n_1 $$ $$ \frac{p_2 V} {R T_2} = n_2 $$

Now, let's say that we want to fuse both boxes and remove bulkhead, so we will have instead 1 box with volume equal $2V$.

What will be temperature $T$ and pressure $p$ of air in the new box? We can use the same law as above to get:

$$ 2 p V = ( n_1 + n_2 ) R T = \left( \frac{p_1 V} {R T_1} + \frac{p_2 V} {R T_2} \right) R T $$,

thus:

$$ p = \frac { T } { 2 } \left( \frac { p_1 } { T_1 } + \frac { p_2 } { T_2 } \right) $$

And there's my question:

If I had heated air after fusion of boxes to temperature $T'$, in other words I had known temperature, I could say what pressure is. Likewise, knowing pressure $p'$ I could say the temperature. But if I just fuse boxes of air and "let it be" (wait very long for air to mix), what the end result should be? Shouldn't (statistically) the pressure and temperature always fix to some certain, unambiguous state? If yes, how I can find them?

I mean that I have one equation $ \quad $ $ p = \frac { T } { 2 } \left( \frac { p_1 } { T_1 } + \frac { p_2 } { T_2 } \right) $, $ \quad $ and two variables: $T, p$, so how can I find both of them? Did I missed something? Or I must to make some more assumptions?

$\endgroup$
2
  • $\begingroup$ You need a second equation that expresses the conservation of energy. $\endgroup$ – Jerry Schirmer Apr 14 '14 at 19:05
  • $\begingroup$ Jerry is right. Energy of the first ideal gas is given by $U_1 = c_v n_1T_1$, where $c_v$ is molar heat capacity at constant volume. Similarly for the second gas. Conservation of energy means that sum of energies of the two gases equals energy of the resulting gas (use the same formula with new $n=n_1+n_2$ and unknown $T$ to be determined). $\endgroup$ – Ján Lalinský Apr 14 '14 at 21:02
0
$\begingroup$

I think you missed the exchange of heat, as both gasses get in equilibrium. This means gas 1 either takes heat from gas 2 if T1 is lower than T2 or loses heat to gas 2 if T1 is higher than T2. This should give you a second equation.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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