The first law of thermodynamics says the change in internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings. Mathematically:

$$\Delta U=Q-W$$

For a system undergoing a quasistatic process, we thus have

$$dU=\delta Q-PdV$$

where $P$ is the pressure of the system and $dV$ is the chance in volume of the system.

The first law is also sometimes written

$$\Delta U=Q+W$$

where this time $W$ is the work done by the surroundings on the system. We thus have

$$dU=\delta Q-P_\text{ext} dV$$

where $P_\text{ext}$ is the pressure exerted on the system by the surroundings.

If the two statements of the fist law are equivalent, then we should have that P_ext=P, but this is only true if the process is reversible.

So what is the error in my reasoning?

After all, by newton's third law, it should hold that the work done by the system on it's surroundings is equal to negative of the work done by the surroundings on the system (since a force applied by the surroundings on system accompanies a force of the same magntidude and in the opposite direction applied by system on surroundings).

  • $\begingroup$ If the system and surroundings are separated by a piston of the mass $m$ and area $S$ then its acceleration $a$ obeys the Netwon's third law $ma=(P-P_\mathrm{ext})S$. If the piston is weightless then $P=P_\mathrm{ext}$, otherwise $P\neq P_\mathrm{ext}$, but you need to take into account change of the piston kinetic energy in the energy balance equations. $\endgroup$ Jun 4, 2017 at 16:13
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/37904/2451 , physics.stackexchange.com/q/39568/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 4, 2017 at 17:32
  • $\begingroup$ Just want to point out that not all thermodynamic systems are PVT systems. $\endgroup$
    – valerio
    Jun 4, 2017 at 20:12

2 Answers 2


Irrespective of whether the process is reversible (quasi-static) or irreversible, the equation for the first law of thermodynamics is always $$\Delta U=Q-\int{P_{ext}dV}$$where $P_{ext}$ is the force per unit are exerted by the gas on its surroundings at the interface between the gas and its surroundings (and by the surroundings on the gas).

In an irreversible process, the behavior of a gas is very different from that at thermodynamic equilibrium, and depends not only on the volume, temperature, and number of moles of the gas but also on the rate at which the volume is changing. In addition, the pressure and temperature of the gas are typically not even uniform spatially within the system. So one cannot use an equation of state (e.g., the ideal gas law) to calculate $P_{ext}$. Instead, $P_{ext}$ must be specified from the external forces that are applied to the gas by its surroundings. These external forces will match those of the gas only at its interface.

On the other hand, for a reversible process, the system passes through a continuous sequence of thermodynamic equilibrium states, and, as a result, the pressure within the system is uniform, and given by the equation of state such as the ideal gas law. So, for a reversible process involving an ideal gas, we can write $$P_{ext,rev}=P_{IG}=\frac{nRT}{V}$$

  • $\begingroup$ That'll clear the doubt. $\endgroup$
    – Mitchell
    Jun 4, 2017 at 19:30
  • $\begingroup$ @chester miller: please identify the error in my reasoning $\endgroup$
    – math_lover
    Jun 4, 2017 at 22:26
  • $\begingroup$ I don't know how else to answer you. The error in your reasoning is, in my judgment, clearly discussed in my response (particularly, the second paragraph). Which part of my response do you feel is difficult to follow? $\endgroup$ Jun 4, 2017 at 23:12

You argument on the last paragraph is correct. If you define the law as $$\delta U = Q + W,$$ then, it should be $$W = P_{\rm ext}dV,$$ which is the energy that the surroundings give to the system. So, your last equation should get a plus sign on the right-hand-side. This implies $\{dV\}_2 = - \{dV\}_1$, because in the second case, the surrounding is expanding when the gas is shrinking. The value of pressure is the same, $P=P_{\rm ext}$, as you argued in your last paragraph.

  • $\begingroup$ In your answer, $W$ has to be equal to $-P_{ext}\Delta V$. If the volume decreases$ (dV = -ve) $the internal energy increases, but according to the signs used in your answer internal energy will decrease. $\endgroup$
    – Mitchell
    Jun 4, 2017 at 15:28
  • $\begingroup$ No. The $P_{\rm ext}$ will be negative, and $dV$ is negative, so the $W$ would be positive. $\endgroup$
    – Jordan He
    Jun 4, 2017 at 15:32
  • $\begingroup$ Pressure is not a vector quantity. In fact, for reversible processes $P_{ext}=P_{internal} \pm dP $ $\endgroup$
    – Mitchell
    Jun 4, 2017 at 15:35
  • $\begingroup$ But it has inside and outside, or negative and positive. $\endgroup$
    – Jordan He
    Jun 4, 2017 at 15:39
  • $\begingroup$ Work can be +ve and -ve. Here, pressure being -ve has no significance. Rather, it is wrong to use -P. $\endgroup$
    – Mitchell
    Jun 4, 2017 at 15:51

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