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1 2 m v ¯ 2 = 3 2 kT . size 12{ { {1} over {2} } m { bar {v}} rSup { size 8{2} } = { {3} over {2} } ital "kT" "." } {}

The kinetic energy of the atoms in a monatomic ideal gas is its only form of internal energy, and so its total internal energy U size 12{U} {} is

U = N 1 2 m v ¯ 2 = 3 2 NkT ,  (monatomic ideal gas), size 12{U=N cdot { {1} over {2} } m { bar {v}} rSup { size 8{2} } = { {3} over {2} } ital "NkT",} {}

where N size 12{N} {} is the number of atoms in the gas. This relationship means that the internal energy of an ideal monatomic gas is constant during an isothermal process—that is, Δ U = 0 size 12{ΔU=0} {} . If the internal energy does not change, then the net heat transfer into the gas must equal the net work done by the gas. That is, because Δ U = Q W = 0 size 12{ΔU=Q - W=0} {} here, Q = W size 12{Q=W} {} . We must have just enough heat transfer to replace the work done. An isothermal process is inherently slow, because heat transfer occurs continuously to keep the gas temperature constant at all times and must be allowed to spread through the gas so that there are no hot or cold regions.

Also shown in [link] (a) is a curve AC for an adiabatic process, defined to be one in which there is no heat transfer—that is, Q = 0 size 12{Q=0} {} . Processes that are nearly adiabatic can be achieved either by using very effective insulation or by performing the process so fast that there is little time for heat transfer. Temperature must decrease during an adiabatic process, since work is done at the expense of internal energy:

U = 3 2 NkT . size 12{U= { {3} over {2} } ital "NkT"} {}

(You might have noted that a gas released into atmospheric pressure from a pressurized cylinder is substantially colder than the gas in the cylinder.) In fact, because Q = 0,  Δ U = W size 12{Q=0, DU"=-"W} {} for an adiabatic process. Lower temperature results in lower pressure along the way, so that curve AC is lower than curve AB, and less work is done. If the path ABCA could be followed by cooling the gas from B to C at constant volume (isochorically), [link] (b), there would be a net work output.

Part a of the figure shows a graph for pressure versus volume. The pressure is along the y axis and the volume is along the x axis. There are two curves. The first curve begins at point A and falls smoothly downward to point B. The graph is shown for an isothermal process. The second curve also begins at point A but falls below the first curve and ends at point C vertically below point B. This graph is shown for an adiabatic process. A line joins point B and C to meet on the X axis. Also a line is drawn from point A to meet the X axis. The area under both the curves is shaded. The graph in figure b is similar to the graph in figure a. Only the directions of the curves are changed. The graph begins from A and moves downward to point B. Then from point B the curve drops vertically downward to C. From point C the graph has a smooth rise back to point A. All directions represented using arrows.
(a) The upper curve is an isothermal process ( Δ T = 0 size 12{ΔT=0} {} ), whereas the lower curve is an adiabatic process ( Q = 0 size 12{Q=0} {} ). Both start from the same point A, but the isothermal process does more work than the adiabatic because heat transfer into the gas takes place to keep its temperature constant. This keeps the pressure higher all along the isothermal path than along the adiabatic path, producing more work. The adiabatic path thus ends up with a lower pressure and temperature at point C, even though the final volume is the same as for the isothermal process. (b) The cycle ABCA produces a net work output.

Reversible processes

Both isothermal and adiabatic processes such as shown in [link] are reversible in principle. A reversible process    is one in which both the system and its environment can return to exactly the states they were in by following the reverse path. The reverse isothermal and adiabatic paths are BA and CA, respectively. Real macroscopic processes are never exactly reversible. In the previous examples, our system is a gas (like that in [link] ), and its environment is the piston, cylinder, and the rest of the universe. If there are any energy-dissipating mechanisms, such as friction or turbulence, then heat transfer to the environment occurs for either direction of the piston. So, for example, if the path BA is followed and there is friction, then the gas will be returned to its original state but the environment will not—it will have been heated in both directions. Reversibility requires the direction of heat transfer to reverse for the reverse path. Since dissipative mechanisms cannot be completely eliminated, real processes cannot be reversible.

Questions & Answers

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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Source:  OpenStax, College physics: physics of california. OpenStax CNX. Sep 30, 2013 Download for free at http://legacy.cnx.org/content/col11577/1.1
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