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In university physics textbook he says :

The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume.

I know that the only contribution to the internal energy comes from the translational kinetic energy (for monatomic ideal gas) according to $$U=K_{trans}=\frac{3}{2} nKT$$

So, obviously, the internal energy $(U)$ depends only on the temperature $(T)$ and the number of moles $(n)$ of the gas. But if someone did work on the gas leading to an increase in pressure and decrease in volume, will this affect the temperature accordingly?

If no, is it because the increase in pressure cancels out with the decrease in volume, and the temperature remains costant according to $$T=\frac{1}{nR}PV$$

If yes, why did he say that the internal energy depends only on the temperature, not on the pressure or the volume ?

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  • $\begingroup$ Molecules in an ideal gas are assumed to have zero interaction with each other. Their only energy, therefore, is kinetic, which is directly related to temperature. $\endgroup$ – Drew Apr 19 at 1:00
  • $\begingroup$ But the temperature itself depends on the pressure and the volume of the gas according to pv=nRT! $\endgroup$ – Karim mohie Apr 19 at 14:45
  • $\begingroup$ You can change the temperature however you like, via changing the volume or pressurizing or whatever. The end effect is that this change in temperature changes the kinetic energy of the molecules, hence the internal energy of the system. You cannot, however, change the kinetic energy of ideal gas molecules without changing their temperature. You might say, "Well look! I can change $P$ and it changes $T$, which changes energy!". Yeah, that's just one way of changing $T$, but the ultimate effect was increasing the kinetic energy $T$ of the molecules, hence the internal energy of the system. $\endgroup$ – Drew Apr 19 at 16:06
  • $\begingroup$ Consider what internal energy is in the first place; it's the energy contained within the system. Ideal gas molecules only have kinetic energy, and temperature is a measure of this average kinetic energy. Therefore, internal energy of an ideal gas is only determined by the average kinetic energy of molecules (i.e., by $T$). Sure there are many ways to change $T$ (changing $P$, $V$, inputting heat, etc.), but that's not the point. $\endgroup$ – Drew Apr 19 at 16:16
  • $\begingroup$ Your argument is analogous to saying that the kinetic energy of a ball thrown by someone depends on some implicit factor, like the strength of the person throwing it. Yeah, that's "technically" true, but more fundamentally it depends on the force at which it was launched. That force may depend on the strength, flexibility, and a bunch of other factors of the person who threw it. $\endgroup$ – Drew Apr 19 at 16:19
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For an ideal gas, you have that $U = \frac{3}{2}nRT$ and also $PV = nRT$, which means that you can write

$$U = \frac{3PV}{2}$$

if you'd like.

It doesn't make sense to say that $U$ is a function of $T$ in no way affected by $P$ and $V$, because (via the ideal gas law) $P,V$, and $T$ are all related to one another. Instead, think of it as the fact that $U$ is determined completely by $T$. If you know $T$, then you know $U$, full stop.

In particular, knowing how $T$ changes tells you immediately how $U$ changes. What happens to $U$ during an isothermal process? Well, if $T$ doesn't change, then $U$ doesn't change. That's it.

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The internal energy of an ideal gas is defined as $ dU = dQ - p dV$:

From Joule's expansion experiment, we see that an ideal gas can expand adiabatically with no change in temperature. Therefore, $d Q =0$ (adiabatically) and $dW=0$ no external work was done. We can conclude that since the pressure and volume of the gas have changed in this process but the internal energy remained unchanged that $U = f(T)$.

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But if someone did work on the gas leading to an increase in pressure and decrease in volume, will this affect the temperature accordingly?

Yes and no.

The answer is yes for a reversible adiabatic ($Q=0$) compression. From the first law

$$\Delta U=Q-W$$

If $Q=0$, $\Delta U=-W$.

And since for an ideal gas $\Delta U=mC_{v}\Delta T$,

$$mC_{v}\Delta T=-W$$

Finally, since work $W$ done on the gas is negative, this means there will be an increase in temperature, as well as pressure, due to the compression.

The answer is no for a reversible isothermal compression, PV=constant, where the heat rejected equals the work done and the change in internal energy is zero,

$$\Delta U=Q-W=0$$

For an ideal gas $$\Delta U=mC_{v}\Delta T=0$$. Pressure goes up when volume decreases and temperature is unaffected (remains constant).

If no, is it because the increase in pressure cancels out with the decrease in volume, and the temperature remains costant according to $$T=\frac{1}{nR}PV$$

As already indicated it is no for an isothermal compression, $Pv$ = constant. From ideal gas law

$$Pv=mRT$$

Therefore $T$ = constant.

If yes, why did he say that the internal energy depends only on the temperature, not on the pressure or the volume ?

The internal energy of an ideal gas consists of only kinetic energy. If work is done on the gas energy is added to the gas. That means the internal kinetic energy has to increase. And since the internal energy of an ideal gas depends only on temperature, that means the temperature has to increase.

Hope this helps.

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For an arbitrary solid, liquid, or gas, it follows from the 1st and 2nd laws of thermodynamics thatthe internal energy per mole is related to the temperature T and specific molar volume V by $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$For an ideal gas, where PV=RT, the term in brackets is zero.

So, in short, if one accepts the 1st and 2nd laws of thermodynamics, the only reason the internal energy of an ideal gas is a function only of temperature is precisely because its equation of state is PV=RT.

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  • $\begingroup$ Does your first equation allow for dissipative effects, or is it only for reversible processes? $\endgroup$ – looksquirrel101 Apr 19 at 12:17
  • $\begingroup$ @looksquirrel101 Internal energy is a function of thermodynamic equilibrium state, and it is independent of any specific process or process path between equilibrium states. $\endgroup$ – Chet Miller Apr 19 at 12:31
  • $\begingroup$ But if we solved this equation (PV=nRT) for the temperature, and substituted by the value of T in K_{trans}, we will get the internal energy as a function of P and V! U=3KPV/2R. Is this equation valid? $\endgroup$ – Karim mohie Apr 19 at 14:53
  • $\begingroup$ It would be valid exclusively for the case of an ideal gas. Is that a problem for you? In reality, it still would be a function only of T. $\endgroup$ – Chet Miller Apr 19 at 15:03
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    $\begingroup$ @looksquirrel101 Yes and yes. $\endgroup$ – Chet Miller Apr 22 at 14:04

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