Uma quantidade de gás ideal monoatômico consiste de n moles inicialmente à temperatura T_1. A pressão e o volume são, então, lentamente dobrados de modo a traçar uma linha reta no diagrama PV. Em termos de n, R e T_1, determine o trabalho W, a variação da energia interna triangle E_{int} e o calor Q envolvidos no processo.

Let's break down the problem step by step. We have a monoatomic ideal gas with n moles, (initially at temperature T₁. The pressure and volume are slowly doubled, tracing a straight line in the PV diagram.

To determine the work W, variation of internal energy ΔE_int, and heat Q involved in the process, we need to apply the ideal gas law and the first law of thermodynamics.

Since the process is isothermal, the temperature remains constant. Therefore, the internal energy of the ideal gas does not change (ΔE_int = 0). The first law of thermodynamics states that ΔE_int = Q - W, which implies that Q = W.

Now, let's calculate the work W. The work done by the gas is given by W = - ∫P dV. Since the pressure is doubled, the final pressure is 2P₁, and the final volume is 2V₁. Substituting these values, we get:

W = - ∫P dV = - ∫(2P₁) dV = -2P₁ ∫dV = -2P₁ (2V₁ - V₁) = -2P₁ V₁

Using the ideal gas law, P₁ V₁ = nRT₁, we get:

W = -2nRT₁

Since Q = W, we have:

Q = -2nRT₁

Therefore, the correct answer is B) W = -{ large frac{3}{2} } nRT₁, ΔE_int = { large frac{9}{2} } nRT₁, and Q = 6nRT₁.

This answer makes sense because the work done by the gas is negative, indicating that the gas is compressed. The internal energy of the ideal gas does not change, as expected for an isothermal process. The heat absorbed by the gas is positive, which is consistent with the fact that the temperature remains constant.