Change in state {for example from liquid to gas} is an isothermal process but the change in internal energy is not zero in this process isn't this contradictory?
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$\begingroup$ i have a test 2morrow so i need the answer fast $\endgroup$– userklsCommented Feb 8, 2015 at 13:22
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$\begingroup$ $dE=TdS-PdV$. I see no contradiction... $\endgroup$– Ryan UngerCommented Feb 8, 2015 at 13:27
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$\begingroup$ so change in internal energy is not a function of change in temperature then why is the change in internal energy taken to be zero in an isothermal process? $\endgroup$– userklsCommented Feb 8, 2015 at 13:35
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$\begingroup$ Are you thinking of $Q=mC_p \Delta T$? $\endgroup$– Ryan UngerCommented Feb 8, 2015 at 13:40
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$\begingroup$ nope i know that is heat its just that we are taking the change in internal energy to be zero for an ideal gas in an isothermal process in my text $\endgroup$– userklsCommented Feb 8, 2015 at 13:43
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3 Answers
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Nope, no contradiction. The law is about energy conservation:
$$\Delta E_{system}=E_{in}-E_{out} \Rightarrow \\ \Delta U = Q-W$$
There is no work $W$ done if we assume constant volume. Internal energy $U$ is changing as you mention. But to vaporize or melt something, even though the temperature is constant you must apply heat $Q$. So $Q$ is also not constant, and: $\Delta U =Q$.
If the material will expand during melting/vaporizing, then there will be work and $W$ is not zero:
$$\Delta U=Q-W$$
Then even more heat $Q$ has to be added since some is used as work for the volume change. But the law of conservation of energy is still true.
Energy input must equal energy output. If it doesn't then the energy of the system must change, since the total sum must be zero (no energy disappears or shows up out of nothing). Your input is heat $Q$. The outgoing energy would be the work, if there is any. Since the input of heat is larger than the outgoing energy (or else the material would not melt), the internal energy rises.
This rise could result in a temperature increase. But since your system is at a point where temperature cannot continue rising without a phase change, the energy must first be used for this phase change. After that, the temperature will continue to rise if you continue heating it up.
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$\begingroup$ How can you say no work is done if it is not stated that the system has constant volume? $\endgroup$ Commented Feb 8, 2015 at 13:41
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$\begingroup$ @0celo7 Good point. Corrected. Thank you. $\endgroup$– SteevenCommented Feb 8, 2015 at 13:43
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$\begingroup$ but isn't work always done in the case of vaporizing or melting something since there is change in volume? $\endgroup$– userklsCommented Feb 8, 2015 at 13:50
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$\begingroup$ @userkls Not in a container e.g. But else, yes. And then $W$ would just not be zero. And therefore you need to apply even more heat to melt (or vaporize) your material, since some is used for work: $\Delta U=Q-W$. Still the law of energy conversion is true. $\endgroup$– SteevenCommented Feb 8, 2015 at 19:07
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Temperature is not a good proxy for internal energy. If anything, you can roughly state that the temperature is the "average energy per degree of freedom" (a result usually called the equipartition theorem), but that still means that energy can flow into or out of a system without disturbing the temperature, if the number of degrees of freedom change.
That is precisely what is happening at a phase transition: for example from solids to liquids, the atoms in a solid which were stuck vibrating about one point suddenly begin to pick up translational freedoms with some of the energy that you're pumping in.
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It may be shown that the energy of an ideal gas is only a function of temperature, $$U=U(T)$$ This is shown in Fermi's book Thermodynamics in section 2.3. I'll update my response if you cannot obtain said text.
answered Feb 8, 2015 at 13:53
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$\begingroup$ yes it is a function of only temperature so if there is no change in temperature then there is no change in energy $\endgroup$– userklsCommented Feb 8, 2015 at 14:02
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$\begingroup$ Yes. It's (simple) calculus: $$dU=\frac{dU}{dT}dT$$ If $dT=0$, then obviously $dU=0$. Does that answer your question? $\endgroup$ Commented Feb 8, 2015 at 14:03
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$\begingroup$ no i got it an ideal gas cannot be liquefied and this relation is only for an ideal gas right so the whole change in state concept is not taken here sorry that was a dumb question $\endgroup$– userklsCommented Feb 8, 2015 at 14:04