One mole of ar initially at 305 k undergoes an adiabatic expansion against a pressure pexternal = 0 from a volume of 8.5 l to a volume of 82.0 l. part a

An adiabatic process is when the system is insulated that no heat is released to the surroundings. For this type of process, we have a derived formula written below:

(T₂/T₁)^C = (V₁/V₂)
where C = Cv/nR

From the complete problem shown in the attached picture, Cv = (3/2)R. Thus,
C= (3/2)/1 mol = 3/2

(T₂/305 K)^(3/2) = (8.5 L/82 L)
Solving for T₂,
T₂ = 67.3 K

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.