A monatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.03 m³ to 0.21 m³. The final pressure of the gas is 60 kPA. The ideal gas constant is R = 8.314 J/mol ∙ K. The change in the internal (thermal) energy of the gas is closest to A) 0.00 kJ. B) 12 kJ. C) 25 kJ. D) -12 kJ. E) -25 kJ.

Question

Quizplus · Accepted Answer

Since the process is isothermal, the temperature remains constant at 300 K. Therefore, ΔU = 0. The first law of thermodynamics states that ΔU = Q - W, where Q is the heat added to or removed from the system and W is the work done by or on the system. Since the process is isothermal and the gas is ideal, we know that Q = W. The work done by the gas is given by the integral W = ∫PdV, where P is the pressure and V is the volume. Thus,

W = ∫PdV = nRT∫dV/V = nRTln(V2/V1)

where n is the number of moles of gas. Using the given data, we find

W = (1 mol)(8.314 J/mol ∙ K)(300 K) ln(0.21/0.03) = 0 J

Therefore, ΔU = Q - W = 0 J. The change in internal energy of the gas is zero.

A Monatomic Ideal Gas Undergoes an Isothermal Expansion at 300