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For a thermodynamics experiment I had to determine the ratio of the volumes of two tanks. The process used was isothermal. I pressurized up one tank and slowly opened the valve between the two tanks and let the pressure equalize. This experiment worked fine.

I was able to derive the necessary equation using the ideal gas law and considering the initial and final states of the two tanks:

$\frac{V_1}{V_2} = \frac{P_2}{P_1}$

Now I'm trying to do the same thing using an isobaric process. Pressurize the tanks up by the same amount and open the valve between them. But I can't seem to derive Charles' Law from the ideal gas equation. This is what I have so far:

Initial state of the two tanks: $PV_1 = M_1RT,\;PV_2 = M_2RT$

Final state of the two tanks: $PV_1 = M_3RT_1,\;PV_2 = M_4RT_2$

I can get to: $\frac{V_1}{V_2} = \frac{m_1}{m_2}$ or $\frac{V_1}{V_2} = \frac{m_3T_1}{m_4T_2}$

I'm aiming for:

$\frac{V_1}{V_2} = \frac{T_1}{T_2}$

Where am I going wrong?

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    $\begingroup$ You got it right - you just need to solve for $V_{2}$ in your equation: $V_{2}/T_{2}= V_1/ T_1$ - which implies $V_2 = ({T_2}/{T_1})V1$. Hence $V_2$ will increase with increasing $T_{2}$ temperature. $\endgroup$ Commented Jul 3, 2019 at 3:48
  • $\begingroup$ If the gases are at the same pressure and same temperature why would you expect something to change? Do you have two different gases (chemicallydifferent) and expect diffusion to happen? $\endgroup$
    – nasu
    Commented Mar 11, 2020 at 14:31
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    $\begingroup$ Charles law as presented in your last equation is usually given for the simplest case of using the same gas or a homogenous mixture of gases. Either way, $M_1=M_2=M_3=M_n$ or is just ignored. $\endgroup$
    – J. Manuel
    Commented Aug 31, 2021 at 13:01

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You must ALSO assume that the numebr of particles is constant, i.e. $M_2=M_3$. You can just call it $M$, or $n$.

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  • $\begingroup$ Using that approach yields T = T1 which can't be correct. However if I do a mass balance on the process I get the required result. $\endgroup$ Commented Nov 20, 2017 at 10:39
  • $\begingroup$ @DarthVader true, but you can do M1+M2=M3+M4 $\endgroup$
    – Kregnach
    Commented Jan 3, 2022 at 22:01

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