Posted by **Ron** on Saturday, April 19, 2008 at 9:14am.

Three moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 430 to 160 K.

(a) What is the work done by (or done to) the gas? Include the algebraic sign.

J

(b) What is the change in the internal energy of the gas? Include the algebraic sign.

J

- Physics -
**Count Iblis**, Saturday, April 19, 2008 at 12:51pm
dE = T dS - PdV

Adiabatic change: Heat absorbed T dS = 0, the change in internal energy is due to work

P dV performed by the gas.

Monoatomic gas:

E = 3/2 N k T

1 mole = 6.022*10^23

k = 1.38*10^(-23)J/K

So, the temperature change gives you know Delta E (which is clearly negative) and that equals minus the work done by the gas (this is positive as work done by the gas comes at the expense of the gas)

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