# Thermodynamics confusion

But doesn't the Internal energy=Potential Energy+Kinetic Energy+...other many energies? So the equation now is Heat supplied=K.E+P.E+....+Work done But doesn't the work done on or by a system is stored as potential energy itself?Then why do we write both P.E and dW in the equation of the first law? Why isn't Heat supplied=Change in internal energy correct?

If this were done in vacuum would there still be work done against something or what?Please help

The First Law refers to a closed (no matter in or out) system and so the word "internal" is all to do with is happen internally in the system and not the surroundings.
Internal energy does not include the kinetic energy of a system due to its motion as a whole or the potential energy of a system due to its position as a whole.
Lifting a container full of gas does increase the potential energy of the gas and the container but it does not increase the internal potential energy of the gas and making a container of gas move faster does increase the kinetic energy of the gas and the container but does not increase the internal kinetic energy of the gas.

In the context of potential energy it is to do with the work done by internal forces (Newton's third law pairs) in changing the (internal) potential energy of the system. So if one changes the separation of interacting atoms within the system the (internal) potential energy of the system will change.
In the context of kinetic energy it is the (internal) kinetic energy due to the random motion of the atoms which is of interest. So changing the average speed of the random motion of atoms will change the (internal) kinetic energy of the system.

When using the First Law the work done on the system is the work done on the system which could change its internal (random motion) kinetic energy and/or its internal potential energy not the work done in lifting the system as a whole or making the system as a whole.

• what are you trying to conclude with 'K.E and P.E of gas and container' and 'Internal energy of the gas'?Aren't they the same thing? – Suraz Basnet Jan 20 '17 at 6:04
• Also,thanks a lot.This cleared my doubt upto higher extent :) – Suraz Basnet Jan 20 '17 at 6:05
• I am trying to differentiate between the pe and KE of the gas and container as a whole and the KE and pe of the gas when individual atoms of the gas are considered. – Farcher Jan 20 '17 at 6:07
• I now know that internal energy doesn't include K.E and P.E of gas and container as whole.But wouldn't the equation then be:dQ=dU+dW+dK.E because the kinetic energy of gas and container as a whole is not incorporated by dW and dU?? – Suraz Basnet Jan 20 '17 at 6:11
• The ke and pe of the gas as a whole is not included in the First Law. So one does not worry about a force which does work accelerating the gas as a whole i.e. accelerates the centre of mass of the gas or lifts the centre of mass of the gas up. One does worry about a force which does work compressing the gas which, with no heat input, would increase the internal energy of the gas as the kinetic energy associated with the random kinetic energy of the atoms. – Farcher Jan 20 '17 at 6:21

You might have seen the simplified energy conservation equation ($U$ is potential and $K$ kinetic energy):

$$U_1+K_1=U_2+K_2$$

It simply says that energy before must equal energy after. It is conserved. But what then if I add energy, for example by doing work? Then you include the added part:

$$U_1+K_1+W=U_2+K_2$$

All energy before plus whatever is added equals the energy after. Makes sense, doesn't it? Now let's just rearrange:

$$W=U_2-U_1+K_2-K_2= \Delta U+\Delta K= \Delta E$$

Work equals the change in the total energy $E$. If heat was added as well, you would just include that, and then you have something that looks veeery much like the first law of thermodynamics.