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One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature $T$. The face of the piston opens into the atmosphere. What is the heat supplied in the process to expand the gas from volume $V_1$ to $V_2$ isothermally ? Friction of the piston against the cylindrical wall is negligibly small.

Now thermodynamics is a common topic to both physics and chemistry and as per my understanding of thermodynamics,I am getting different results for the heat that will need to be supplied.

According to physics :$$∆Q =∆U+∆W$$

And here the Work done is the Work done by the gas. That is :

$$W = \int PdV=\int \frac{RT}{V}dV = RT( \ln V_2)-RT( \ln V_1)$$ and as for Isothermal process: $$∆U =0 , ∆Q =RT( \ln V_2)-RT( \ln V_1)$$

Chemistry on the other hand says: $$∆U =∆Q +∆W $$ and defines it as the Work done on the gas and this will give: $$W = -P_{atm}(V_2-V_1) $$ and as for isothermal process:$$ ∆U =0 , ∆Q =P_{atm}(V_2-V_1)$$

I am also uncertain that maybe neither of these is the correct expression and the network is actually in that work from both these forces and we have to take the sum of these to yield :$$∆Q =RT( \ln V_2)-RT( \ln V_1)-P_{atm}(V_2-V_1)$$

Now the value of $Q$ should be unique because in the real world when we perform an experiment, only one value of $Q$ would be supplied.

So which one of these is correct and where am I lacking in my understanding of thermodynamics from a physics standpoint ??

Now as this question links both physics and chemistry, I want to post it on both sites (I hope it's fine) and the link to it is here

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  • $\begingroup$ Related: physics.stackexchange.com/q/37904/2451 , physics.stackexchange.com/q/39568/2451 and links therein. $\endgroup$ – Qmechanic May 23 at 15:49
  • $\begingroup$ The physics/engineering version is $\Delta U=Q-W$. The Chemistry version is $\Delta U=Q+W$ Both versions are valid and give the same magnitude of work done. The difference is the sign of the work is positive for the physics/engineering version if the work is done by the system (expansion) and the sign is negative for the chemistry version. Both result in the logical decrease in internal energy. Just need to be consistent. $\endgroup$ – Bob D May 23 at 15:54
  • $\begingroup$ But here which of the expression for W is correct ?? @ Bob D $\endgroup$ – RandomAspirant May 23 at 16:44
  • $\begingroup$ How can the temperature be constant if the gas is made to expand at constant pressure? $\endgroup$ – Chet Miller May 23 at 16:53
  • $\begingroup$ @Chet Miller , if I have a perfectly conducting vessel and it's in an ice bath, then the temperature will remain constant. $\endgroup$ – RandomAspirant May 23 at 18:01
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The physics and chemistry examples you gave describe two different processes. In the physics example, the gas is subjected to an isothermal reversible (quasi-static) expansion, where the gas pressure and the external pressure decrease very gradually. For this case, using the ideal gas law to determine both the gas pressure and the external pressure are valid.

However, the chemistry example is much different. Here, the gas is subjected to an "isothermal" irreversible (non-quasi-static) expansion, where the external pressure is suddenly dropped from its initial value and then held at that value until the system re-equilibrates. In such a case, the ideal gas law can not be used to determine the gas pressure because the ideal gas law applies only to thermodynamic equilibrium situations (or to reversible processes where the gas passes through a continuous sequence of thermodynamic equilibrium states). For an irreversible (non-quasi-static) process, the gas passes through sequence of non-equilibrium states, and the ideal gas law is not valid for this. In addition, even though the external boundary of the gas is held at a constant temperature, during this irreversible deformation, the interior temperature of the gas varies with spatial location. So, even though we call the process "isothermal," it is really only isothermal for a tiny fraction of the gas in contact with the boundary. Of course, in the end, the gas temperature throughout will be back to the wall value when the gas reaches thermodynamic equilibrium again.

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  • $\begingroup$ So exactly which work expression is correct ?? I believe, then it should be the chemistry one ?? That is +P∆V ?? $\endgroup$ – RandomAspirant May 24 at 5:57
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    $\begingroup$ If $P_{ext}$ represents the force per unit area exerted by the gas on the inside face of the piston (and, by Newton's 3rd law, also the force per unit area exerted by the inside face of the piston on the gas), then the work done by the gas on the piston is $$W_{by\ gas}=\int{P_{ext}dV}$$ and the work done by the piston on the gas is $$W_{by\ piston}=-\int{P_{ext}dV}$$. **So both are right**. The 1st law can be written as $$\Delta U=Q-W_{by\ gas}=Q+W_{by\ piston}$$ $\endgroup$ – Chet Miller May 24 at 12:11
  • $\begingroup$ Oh okay , Thanks a lot !! I think now am understanding what exactly this is. $\endgroup$ – RandomAspirant May 24 at 16:30
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The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious. The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change. If the volume does change, then either the pressure must change or the temperature must change.

In the "chemistry solution" you presented, the pressure is taken as constant (atmospheric). So the temperature must change. But if the temperature changes, then $\Delta U$ is not zero (as they assume). So the solution is incorrect. The correct solution, based on the 1st law, is $$\Delta U=nC_v\Delta T=Q-P_{atm}(V_2-V_1)$$or equivalently, $$\Delta H=\Delta U+P_{atm}(V_2-V_1)=nC_p(T_2-T_1)=Q$$Also, since $$n=\frac{P_{atm}V_1}{RT_1}$$we have $$\frac{T_2}{T_1}=1+\frac{QR}{C_pP_{atm}V_1}$$

In the "physics solution" you presented, they make the opposite assumption that, rather than the pressure remaining constant, the temperature remains constant. For the temperature remaining constant, the solution you presented is correct. But the final pressure is: $$P_2=P_{atm}\frac{V_1}{V_2}$$

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Both versions of the first law are valid. In engineering the version is Δ𝑈=𝑄−𝑊. The Chemistry version is Δ𝑈=𝑄+𝑊. In physics it varies. Both versions give the same magnitude of work done. The difference is the sign of the work is assigned a positive value in the version if the work is done by the system (expansion), and the sign is negative for the second version. Both result in the logical decrease in internal energy as the system spends energy doing work. The important thing is to be consistent.

REVERSIBLE ISOTHERMAL WORK;

For both versions, the magnitude of work done by the system for a reversible isothermal expansion is

$$W=RTln\frac{V_2}{V_1}$$

For the chemistry version it is given a negative sign, for the engineering/physics version a positive sign.

For either version, the work done in the problem as stated can not be reversible isothermal work. For a reversible isothermal expansion the product of pressure and volume is a constant. If the volume increases the pressure must decrease. So the reversible process can't involve a constant external pressure ($P_{atm}$) as you've written. It would have to be an irreversible isothermal process where only the temperature at the interface between the system and atmosphere is constant. The temperature within the gas is undefined because of temperature and pressure gradients.

Hope this helps.

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  • $\begingroup$ One minor quibble. It gives the wrong impression to say that $Q-W$ is the physics version (I can't speak for engineering). Plenty of physics texts use $Q+W$. $\endgroup$ – garyp May 25 at 13:17
  • $\begingroup$ @garyp True in physics it is sometimes +W. I haven't done a formal survey, but more often than not it seems to me it is -W. Anyway, I will revise so as not to seem so absolute. As far as engineering is concerned, the NCEES (National Council of Examiners for Engineering and Surveying) reference handbook for the Fundamentals in Engineering (FE) exam shows it as -W. $\endgroup$ – Bob D May 25 at 13:45
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Both the equations are right,the second equation is just a special case of the first equation(when pressure is a constant, we note

$\frac{RT}{V}$

is also constant and it comes outside the Integral sign which leaves us with the equation -A result from the ideal gas equation

$P(V_2-V_1)$).

Hope this helps

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  • $\begingroup$ Why do you think is the first one correct over the other ? $\endgroup$ – RandomAspirant May 23 at 15:48
  • $\begingroup$ oops I meant the first one is the more general version.Both are correct, but the second one is a consequence of the former. $\endgroup$ – Monocerotis May 23 at 15:51

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