# Work done by gas in isothermal process [closed]

A $$20\ \mathrm l$$ steel cylinder containing $$\ce{N2(g)}$$ at $$20\ \mathrm{atm}$$ and $$300\ \mathrm K$$ is kept in a big jar containing air at $$1\ \mathrm{atm}$$ and $$300\ \mathrm K$$ and sealed. $$\ce{N2(g)}$$ starts leaking into the jar but reversibly and isothermally. When both mechanical and thermal equilibria are stabilized, pressure in the jar was $$1.19\ \mathrm{atm}$$. Determine work done by $$\ce{N2(g)}$$.

Since there is mechanical equilibrium, the pressures should be equal to $$1.19\ \mathrm{atm}$$ inside and outside the cylinder with nitrogen. And since temperatures are equal too, should I use this formula for work? $$W=-nRT\ln\frac{V_{2}}{V_{1}}$$

• This is a very poorly posed problem in my judgment. The process you described does not seem to be reversible. Did the problem mean to say "irreversible"? Even so, once the N2 leaks into the air, it mixes with the air. So how does it do work on the air? – Chet Miller Jan 27 '19 at 14:16
• @ChesterMiller I think they mean to say the work done against the pressure of the air around it, and that pressure at equilibrium, of N2 would be 0.19atm by partial pressure law. Then work will be -P∆V – user638473 Jan 27 '19 at 14:20
• So you are assuming that the N2 does not undergo mixing by diffusion into the air once it escapes? If this is the calculation, then the first step is to determine the number of moles of air in the jar, such that the final pressure is 1.19 atm in both the cylinder and the jar. – Chet Miller Jan 27 '19 at 14:25
• Yes, and also I assume that the air does not enter the cylinder containing N2 – user638473 Jan 27 '19 at 14:31

The compression of the air outside the cylinder by the $$\ce{N2}$$ escaping from the cylinder takes place reversibly (as far as the air is concerned), but the process of $$\ce{N2}$$ leaking out of the cylinder through a valve or small orifice is definitely irreversible because of the Joule–Thompson throttling that takes place as the $$\ce{N2}$$ flows from a high pressure inside the cylinder to a much lower pressure outside the cylinder. So to calculate the work done by the $$\ce{N2}$$, one cannot treat the expansion of the nitrogen as reversible. Instead, one must focus on the reversible work required to compress the air.

Assuming that the $$\ce{N2}$$ does not diffuse into the air (i.e., the boundary between the air and the $$\ce{N2}$$ remains well-defined), the final volume occupied by the $$\ce{N2}$$ is $$11 × 20/1.19 = 184.9$$ liters. That means that the volume of $$\ce{N2}$$ increased by $$184.9 - 20 = 164.9$$ liters. This is the volume of $$\ce{N2}$$ outside the cylinder (in the jar) in the final state of the system. If $$V_0$$ represents the volume in the jar outside the cylinder (and also the initial volume of air), the final volume of air is $$V_0 - 164.9$$. So, the ratio of the initial volume of air to the final volume of air is equal to the pressure ratio of the air, and given by:

$$\frac{V_0}{V_0 - 164.9} = 1.19$$

Solving for $$V_0$$ then yields $$V_0 = \pu{1033 L}$$.

The work done by the air on the $$\ce{N2}$$ is given by

\begin{align} W_\mathrm{a} &= n_\mathrm{a}RT\ln{\left(\frac{1033 - 164.9}{1033}\right)} \\ &= P_0V_0\ln{\left(\frac{1033-164.9}{1033}\right)} \\ &= 1\cdot 1033\ln{\left(\frac{1033-164.9}{1033}\right)} \\ &= \pu{-180 L atm} \end{align}

This is equal in magnitude and opposite in sign to the work done by the $$\ce{N2}$$ on the air. So the work done by the $$\ce{N2}$$ on the air is 180 liter-atm.