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Consider a system of ideal gas in a container with piston and the isobaric expansion of gas takes place.
As the process is isobaric, so initially the pressure of gas is equal to atmospheric pressure as the piston remains in equilibrium over the container. Then, first we increase the temperature and as a result pressure of gas increases and the piston rises till the pressure reduces to atmospheric pressure. So at initial and final state pressure of gas is constant thus the process is isobaric.
So, work done by gas should be
$dW=F_{applied\;by\;gas}(x).dx$
$dW=P_{applied\;by\;gas}(x)Adx$
$s.t.\;P_{applied\;by\;gas}(x_{initial\;state})=P_{applied\;by\;gas}(x_{final\;state})=P_{atm}$
So, total work done, $W=\int_{x_{initial}}^{x_{final}}P(x)Adx$
So, we should know pressure as a function of $x$ to calculate the work done by gas.

My question is that in books, they directly use the argument that work done by gas is dependent on $P_{external}$ and as it is constant, they give the expression of work as $dW=-P_{external}Adx$.
I don't understand how they do that and use external pressure in the expression of work done.
I think we should take into account pressure of gas as a function of $x$ to find the work done by gas.
Please help me in clarifying the doubt.

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    $\begingroup$ Isobaric means with constant external pressure. Then it depends if the process is reversible ( always considered in equilibrium ) or not. For the former, internal and external pressures are equal and constant. For more, see Reversible process on wikipedia. $\endgroup$
    – Poutnik
    Dec 9 '20 at 11:09
  • $\begingroup$ Consider that for reversible isobaric expansion $p_\mathrm{int} = p_\mathrm{ext} + \mathrm{d}p$ $\endgroup$
    – Poutnik
    Dec 9 '20 at 11:18
  • $\begingroup$ Thanks for the reply. But if the process is irreversible then I think we have to take into account the pressure of gas as a function of x. $\endgroup$
    – Manu
    Dec 9 '20 at 11:19
  • $\begingroup$ Because then $p_{int}=p_{ext}+\Delta p$ and $\Delta p$ can't be neglected $\endgroup$
    – Manu
    Dec 9 '20 at 11:21
  • $\begingroup$ For example at initial and final state, pressure of gas is same as external pressure andbetween the states it varies as explained in my question. But, in high school chemistry book, it is written that work done for doth reversible and irreversible process is -$P_{ex}\Delta V$. I don't understand that. $\endgroup$
    – Manu
    Dec 9 '20 at 11:26
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At the interface between the gas and its surroundings, by Newton's 3rd law, the pressure exerted by the gas on its surroundings is equal to the pressure exerted by the surroundings on the gas. So we can use either. For a reversible expansion, the gas pressure is given by the ideal gas law. But the ideal gas law only applies at thermodynamic equilibrium and during a reversible process (which consists of a continuous sequence of thermodynamics equilibrium states). For an irreversible expansion, the ideal gas law doesn't work (i.e., gives the wrong answer), and we can't use it to get the gas pressure at its interface with the surroundings (where the work is being done). But, if we know the surroundings pressure during the irreversible expansion, we can still calculate the work (because, at the interface between the gas and surroundings), the two will match. So using the external pressure is always a valid approach to getting the work.

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  • $\begingroup$ I think that Newton's third law is not being applied in the way you suggest. If gas particles collide with the surface of piston with force F or pressure P, then the equal and opposite force will be exerted by the piston not by the surroundings. $\endgroup$
    – Manu
    Dec 9 '20 at 13:36
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    $\begingroup$ The piston is part of the surroundings. The inside face of the piston is the interface between the gas and its surroundings; this is where the action-reaction pair is situated. And, if the piston is massless, it doesn't matter whether the piston is included within the system, or within the surroundings. $\endgroup$ Dec 9 '20 at 13:46

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