Question

A piston containing 3.00 mol of Ar undergoes a process taking it from step one to two below. In all cases, P_ext = P_int. C_p(Ar) = 5/2 R

Ar(30.0 L³, 3.00 atm) = Ar(90.0 L³, 1.00 atm).

Calculate w, q, and delta U. If equal to 0, explain why.

Answer 1

Answer.

Given that,

Initial Pressure (P₁) = 3 atm
Final Pressure (P₂) = 1 atm
Initial Volume (V₁) = 30 dm³ = 30 L
Initial Volume (V₂) = 90 dm³ = 90 L

Since P1V1 = P2V2, the given process is isothermal process.

For an isothermal process, since there is no temperature change.

Hence, according to first law of thermodynamics ,

Q = W.

Now,

W= work done = (for an isothermal process)

Using ideal gas equation pV = nRT,

W = = -90*1 ln(90/30) = -90 ln(3) L-atm = -101.325*90*ln(3) J
(since 1 L-atm = 101.325 L-atm)

W = -10018.5201 J = -10.02 kJ

Q = W = -10.02 kJ.