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A monatomic ideal gas expands slowly to twice its original volume, doing 370 j of work in the process. part a find the heat added to the gas if the process is isothermal.

User Aaayumi
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The first law of thermodynamics says that the variation of internal energy
\Delta U of a gas is equal to the amount of heat Q supplied to the gas minus the work W done by the gas:

\Delta U = Q-W

The variation of internal energy of a gas is:

\Delta U = (3)/(2) n R \Delta T
As it can be seen, it depends only on the variation of temperature
\Delta T. Since for an isothermal process
\Delta T=0, then
\Delta U=0. This means that the first law of thermodynamics becomes

Q=W
and since the work done is 370 J, then the amount of heat is also 370 J:
Q=370 J.
User Bastianneu
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