This is a problem from Heat and thermodynamics by zemansky. A thin-walled metal container of volume $V$ contains a gas at high pressure.Connected to the container is a capillary tube and a stopcock.When the stopcock is opened slightly,the gas leaks slowly into a cylinder equipped with a non-leaking,friction less piston, where the pressure remains constant at atmospheric pressure,$P_0$.
a)Show that, after much gas has leaked out,an amount of work $W=-P_0(V_0-V)$ has been done where $V_0$ is the volume of gas at atmospheric pressure, $P_0$ and temperature.
b)How much work would be done if the gas leaked directly into the atmosphere?
The first part is easy to prove. The process is quasi-static therefore we integrate $dW=-PdV$ from $V$ to $V_0$. The second part is confusing me. I think it should be the same amount of work, since the piston was frictionless, the gas would just have to work against the air molecules just like the second scenario. But how can there be no difference? the volume with the piston is finite whereas when leaked directly into the atmosphere it has infinite space to occupy. What is the right explanation?
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$\begingroup$ For the most part, the gas escaping from the capillary retains its integrity, just as if there were an invisible membrane separating it from the rest of the atmosphere. $\endgroup$– Chet MillerCommented Jun 6, 2020 at 22:33
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$\begingroup$ why would it do that? $\endgroup$– 1500kook12Commented Jun 7, 2020 at 11:27
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$\begingroup$ What do you think would happen? $\endgroup$– Chet MillerCommented Jun 7, 2020 at 12:09
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$\begingroup$ intuitively i feel it should dissolve into the atmosphere...as soon as it leaks out of the stopcock $\endgroup$– 1500kook12Commented Jun 7, 2020 at 12:29
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$\begingroup$ Only if the flow is rapid and turbulent. But, out of a capillary and stopcock, it will be more as I described. $\endgroup$– Chet MillerCommented Jun 7, 2020 at 12:56
1 Answer
The later will be sudden adiabatic expansion...intermediate states are non-equlibrium (can not be integrated) but from the knowledge of the final & initial state one can calculate W..think in two steps; 1)you can think as if the gas becomes in atmospheric pressure consistently with its eqn of state (as if inside a very very thin walled [ultimately taking the density of material of this wall to tend to zero in appropriate sense] expanding baloon) & 2)once it has equal pressure of the atmosphere the molecules get scattered (now no pressure difference & no work)..so we need to get know w for the 1st step...from just initial & final state of this non-equl. process: ∆u + W = 0 (Roy Gupta book convention or a negative sign as per zeemansky convention))..for ideal gas ∆u = n c_v ∆T =& can be connected to P_0 (atmospheric pressure), v_0 , v etc parameters as in zeemansky book problem..
