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