Physics
posted by Ron on .
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

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)