I have this problem from Resnick's Physics, 5th ed, problem 31-15, sorry I only have the Spanish version (i'll translate):

A resistor R=550 ohm, connected to an external battery, is inside an adiabatic cylinder with a frictionless piston containing an ideal gas. A current i=240 mA flows through the resistor. ¿At what velocity must the piston (mass=11.8 kg) move upwards, in order to keep the gas temperature unchanged?

I came across the solution manual in English and it gives this:

solution .

I think there are many wrong things here, like, did they neglect atmospheric pressure? OK, even if we assume there's a vacuum outside, still. Let me break down their resoning:

-First Law: $\frac{dU}{dt}=\frac{dQ}{dt}-\frac{dW}{dt}$
-the electrical power is all dissipated in the resistor and received by the gas as heat. It is constant. -The expansion work, regardless of quasi-static process or not, is the lifting of the piston. There would be extra work if there was atmospheric pressure and if the piston was accelerating.
-U does no vary, and neither does the velocity, so there is no extra work due to acceleration.

$0=i^2R-mgv$, solve for v.

What reasoning is been done in these equations? If it was possible for $\dot Q$, $\dot U$ and v to be simultaneously constant, and I canmathematically find the solution for v, then, the process is possible.

BUT constant v implies constant pressure ($P=\frac{mg}{S}$). And a process in an ideal gas cannot be isobaric and isothermal at the same time, right?

I want to ask you guys to try to explain to me if the book's solution is possible and correct, or to solve the problem correctly or rephrase it, like, either T or v aren't constant or there's something wrong in assuming the heat the gas receives is constant. What would be the "real" behavior of this (ideal) system if it was given, like, some initial temperature and speed? Would it reach a steady state? Or is there a function v(T,V) or v(U,V) or U(v,V) we can calculate? Thanks!

  • $\begingroup$ I figured I could formulate the problem without assuming U or v constants, even with atmospheric pressure if you like, and the equation is: $\frac{dU}{dt}=i^2R-(mg+P_{atm} S)v-M\frac{dv}{dt}v$ with M the mass of the piston. Now what do I do with that? $\endgroup$
    – dronkit
    Aug 7, 2018 at 2:33
  • $\begingroup$ my bad, both m's should be lowercase. $\endgroup$
    – dronkit
    Aug 7, 2018 at 2:44
  • $\begingroup$ The answer I presented assumes that the piston is moving at constant velocity, and demonstrates that your criticisms are well-founded. $\endgroup$ Aug 7, 2018 at 2:53
  • $\begingroup$ I continue to maintain, with confidence, that your original assessment of this was correct. $\endgroup$ Aug 8, 2018 at 0:27
  • $\begingroup$ the only thing I haven't questioned is the assesment that constant velocity implies constant pressure. It sure does if the process is quasi static, butI have no idea what could happen if it isn't. Can it be known? Or, does it even make seense to argue "the process may be not quasi static so it may be not isbaric cause we can't say there is a defined P" but at the same time mantain it's isothermal if we can't tell even if there is a defined T? $\endgroup$
    – dronkit
    Aug 8, 2018 at 1:37

3 Answers 3


Let’s try this again using a different approach, one that specifically responds to question “… try and explain to me if the book’s solution is possible and correct…?” The following is presented for your consideration.

Fig 1 below describes an apparatus (whether or not it is practical is another question). The system is the gas and the resistor. The adiabatic cylinder walls and piston are the boundary. Everything else constitutes the surroundings. The piston supports a weight, which rests on top of the cylinder. All surfaces moving relative to one another are frictionless. A perfect seal is between the piston shaft and opening of the top of the cylinder. Atmospheric air surrounds the weight. Before the heater is turned on, the system is in thermal and mechanical equilibrium. Ideal gas behavior is assumed throughout the process.

Refer now to Fig 2. In order for PxV of the gas to be constant when the heater is turned on, an external agent is needed to apply an upward force to the weight that varies with the height of the piston as shown. Consequently, this is not a constant pressure process. Since the boundary is adiabatic, no heat crosses the boundary and Q=0 in the first law equation. The work done on the gas is electrical work that crosses the defined boundary. (Within the system there is heat transfer from the heater to the surrounding gas. See comments below). If you do the necessary integrations you’ll find the work done by the gas on the weight plus the work done by the external agent on the weight will by $mg \Delta h$ as you would expect.

With this arrangement it would be possible in theory to have a reversible isothermal process if it were not for the fixed electrical power source. Therefore, the process will necessarily be irreversible.

Initial conditions:

$$h = h_o$$ $$F_{gas} = mg$$ $$T = T_1$$

When the heater is initially switched on a differential amount of electrical work done is given by:

$$-I^2R dt$$

During that time the gas does a differential amount of boundary work given by:

$$+mg dh$$

In order for no change in internal energy: $$I^2R dt = mg dh $$


$$\frac{dh}{dt} = \frac{I^2R}{mg}$$

Therefore the book answer is the instantaneous velocity at the start of the process. However, the velocity will not be constant. In needs to increase linearly with height in order to accommodate the fixed rate of electrical work crossing the boundary. For example, when the volume doubles the velocity doubles. In order for the velocity to be constant, we would need an electrical power input that varies inversely with height (volume).

The following are additional comments regarding practicality, temperature, atmospheric air, and representation of the electrical energy input as either Q or W.

Practicality- Fig 1 is not intended to represent a practical apparatus for conducting the process. For example, the distribution of the resistance is limited to the initial boundary. Resistors don’t immediately reach max temperature upon switching on as they are made of material with finite specific heats. All contacting surfaces are subject to friction. The type of external agent that provides the height varying external force is not specified, etc., etc.

Temperature- Since this process is not quasi static there will be, as Chester Miller points out, spatial temperature variations throughout the gas during the process. However, if the internal energy does not change, the average translational kinetic energy of the gas molecules will remain the same. Problem is there is no way to measure it while the heater is on. In order to be able to make temperature measurements, thermal equilibrium is necessary. One approach is to cycle the heater on and off, with short on times and long off times sufficient for the system to reach thermal and mechanical equilibrium. At each equilibrium state the temperature can be measured and should theoretically be the same. In terms of a PV graph we would have a series of equilibrium points, with the process being undefined between points.

Atmospheric Air- For the apparatus shown, the weight is surrounded by atmospheric air. Consequently, it plays no role in the process (this is what I had in mind when I said it didn’t necessarily matter).

Electrical Energy Input as Either Q or W- In this example, the electrical input is considered work transfer (W) crossing the boundary, which is typical in treatments like this. In the presentation of the question and Chester Miller’s analysis, the electrical input is represented as Q. That necessitates a different boundary as shown in Fig 3, in which the heater becomes part of the surroundings. The net effect is, of course, the same.

Hope this helps.

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  • $\begingroup$ thanks! You've made it useful and solved it in great detail. Indeed an external force which makes pressure non-constant (not mentioned in the original) is the only way you can have speed and temperature constant. Ironically I saw the same idea in another spanish forum, but your redaction and solution is most complete and clear. Obviously this doesn't change my opinion that the original problem is flawed. $\endgroup$
    – dronkit
    Sep 1, 2018 at 22:34

Maybe I’m missing something but I don’t see anything wrong with the problem. In order for the temperature not to change, the internal energy cannot change. The rate at which energy (electrical energy) is entering the gas is $i^2R$. The rate at which the gas does work on the piston moving it upward at constant velocity $v$ is $mgv$. It is essentially the rate at which we are increasing the gravitational potential energy of the piston. In order for no change in internal energy they must be equal.

$i^2R = mgv$

$v = i^2R/mg$

As far as the outside pressure is concerned, I’m not sure it matters. Does atmospheric pressure matter in calculating the work done in lifting a mass? The solution implies a constant pressure process which in this case is simply the pressure that the weight of the piston applies to the gas.

  • $\begingroup$ but like I said, how can the process be isobaric and isothermal at the same time? $\endgroup$
    – dronkit
    Aug 7, 2018 at 23:29
  • $\begingroup$ this is secondary, but responding to the atmosphere question, there is a work done against the atmospheric pressure, $P_{atm} \Delta V$, or a "compression of the atmosphere" $\endgroup$
    – dronkit
    Aug 8, 2018 at 1:34
  • $\begingroup$ Regarding your first comment, I don't believe the book problem covers a process that can actually happen. Thats the reason for all the comments. BUT if you assume constant temperature and treat the problem using only the first law without regard to what process is actually occurring, then book solution would be correct. $\endgroup$
    – Bob D
    Aug 8, 2018 at 19:59
  • $\begingroup$ Regarding the second comment, the reason I didn't feel the outside pressure mattered is the problem only requires calculation in the rate of change in potential energy. Let's say atmosphere is present and everything is in equilibrium before turning on the switch. If you then remove the atmosphere, the gas will expand and the piston will rise. So all if affects is the initial height of the piston before the start of the process. $\endgroup$
    – Bob D
    Aug 8, 2018 at 20:04
  • $\begingroup$ I need to warn you to be careful with the atmospheric pressure issue. In thermodynamics you should always keep it in mind. In a pure mechanics problem, i.e., lifting a rock, atmospheric pressure doesnt' matter cause you have it acting below the rock as well as above the rock. In hydrostatics you almost always forget about it because it acts on both sides of walls, objects, etc., or because you work with relative pressures and Patm=0 in that reference frame. Which doesn't mean you are neglecting it. $\endgroup$
    – dronkit
    Aug 17, 2018 at 0:04

In my judgment, your criticisms are totally well founded. In this constant pressure process (assuming, as you said, that the external region is vacuum), the first law applied to an ideal gas tells us that $$\frac{dH}{dt}=nC_p\frac{dT}{dt}=\dot{Q}$$ This clearly means that the temperature is rising, and so is the internal energy. So $$\frac{dU}{dt}=nC_v\frac{dT}{dt}=\frac{C_v}{C_p}\dot{Q}$$ From the first law, we have $$\frac{dU}{dt}=\dot{Q}-\dot{W}$$Therefore,$$\dot{W}=\frac{(C_p-C_v)}{C_p}\dot{Q}=\frac{(\gamma-1)}{\gamma}\dot{Q}$$ or $$mgv=\frac{(\gamma-1)}{\gamma}i^2R$$ So, the temperature is not constant, and the velocity is less than that given in the "official solution" by a factor of $(\gamma-1)/\gamma$. For an ideal diatomic gas, $\gamma = 1.4$. So the factor is equal to 0.286


The only way that this problem can realistically be analyzed within the framework of thermodynamics is if we assume that the rate of heat addition and the rate of volume change are slow enough for the process can be considered quasi static (both thermally and mechanically). Otherwise, we would have to consider that there would be significant spatial temperature-, pressure-, and gas velocity variations within the cylinder, viscous stresses within the gas would contribute to the system response, and we would have to solve the partial differential equations of fluid mechanics and heat transfer (Navier-Stokes equations, differential thermal energy balance) within the cylinder to describe the behavior. So, in our analysis, we make the quasi static approximation.

Constant Pressure: Using a free body diagram of the piston, if we apply Newton's 2nd law to its motion at constant velocity, we obtain: $$PA=mg$$ So the gas pressure is constant. And the rate at which the gas does work on its surroundings is just $$\dot{W}=P\frac{dV}{dt}=mgv=nR\frac{dT}{dt}$$Note that, from the ideal gas law, we have no choice but to say that the temperature is increasing with time (if the volume is increasing and the piston is moving upward).

Note that we haven't said anything about how the piston achieved this constant velocity. Of course, it could be established initially by applying an impulsive force at time zero such that the initial velocity just matches that consistent with the imposed constant rate of heat addition (where the heater is also switched on at time zero).

Constant Temperature: If the heating is not quasi static, then there will be significant gas spatial temperature variations within this cylinder, with the highest gas temperatures near the heater and the lowest away from the heater. However, in our quasi static limit, these spatial temperature variations become negligible. Unfortunately, there is still the issue of the time variation of the temperature. There is no way of getting around the fact that, for an ideal gas, if the pressure is constant and the volume is increasing with time, the temperature must also be increasing with time. So the condition in the problem statement that gas temperature is constant in time cannot be satisfied. This brings us back to the analysis I presented above, in which the required rate of temperature rise is taken into account in determining the steady velocity of the piston. Note that the predicted velocity of the piston in this case is much lower than that based on the inconsistent problem statement; and it thus brings us closer to the ideal quasi static limit.

  • $\begingroup$ @BobJacobsen: > You're assuming a different physical situation[...] it's OK, below the original problem I asked if someone can rephrase it to be solvable and more useful $\endgroup$
    – dronkit
    Aug 8, 2018 at 1:48
  • $\begingroup$ @BobD the pressure is constant if the velocity is constant and low enough for the process to be quasi-static $\endgroup$
    – dronkit
    Aug 8, 2018 at 2:01
  • 1
    $\begingroup$ @ChesterMiller: "Please tell us how the pressure and temperature of an ideal gas can remain constant if its volume changes": you've exposed the contradiction in the problem even better than myself. $\endgroup$
    – dronkit
    Aug 8, 2018 at 2:03
  • $\begingroup$ I deleted a comment discussion which appears to have concluded. $\endgroup$
    – David Z
    Aug 8, 2018 at 6:31

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