With the isobaric expansion of argon taken at normal atmospheric pressure, its internal energy increased by 300 J.

With the isobaric expansion of argon taken at normal atmospheric pressure, its internal energy increased by 300 J. Determine the change in the volume of argon.

Internal energy of ideal gas:
U = (i / 2) * p * V.
U1 = (i / 2) * p * V1 is the energy of argon before expansion, where i is the number of degrees of freedom of argon (i = 3), p is the argon pressure (for n.o. p = 10 ^ 5 Pa), V1 is initial volume of argon.
U2 = (i / 2) * p * V2 – energy of argon after expansion, V2 – final volume of argon
Internal energy change: ∆U = U2 – U1 = 300 J.
∆U = (i / 2) * p * V2 – (i / 2) * p * V1 = (i / 2) * p * ∆V.
∆V = ∆U / (i / 2) * p = 300 / (3/2) * 10 ^ 5 = 0.002 m ^ 3 = 2 dm ^ 3 = 2 l.
Answer: The change in the volume of argon is 2 liters.