The figure (not to scale)shows a pV diagram for 1.8 g of helium gas (He)that undergoes the process 1 → 2 → 3.Find the value of V₃.The ideal gas constant is R = 8.314 J/mol ∙ K = 0.0821 L ∙ atm/mol ∙ K,and the atomic weight of helium is 4.0 g/mol.

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Using the ideal gas law, we can find the initial volume and temperature of the helium gas at point 1: PV = nRT V = (nRT)/P = (0.0018 kg)/(4.0 g/mol) * (8.314 J/mol*K) / (101325 Pa/atm) * (1 atm/101325 Pa) = 0.048 L The temperature is not given, so we cannot find the value of U1. Next, we can find the work done by the gas during process 1 → 2: W = ∫PdV = (1 atm)(0.018 L - 0.048 L) = -0.03 L*atm = -30 J Since no heat is added or removed during this process (ΔQ = 0), we know that ΔU = W, so U2 = U1 + W = U1 - 30 J. For process 2 → 3, we can use the fact that the gas is undergoing an isothermal process (constant temperature), so we have: PV = nRT P1V1 = P2V2 V2 = V1P1/P2 = (0.048 L)(1 atm)/(4 atm) = 0.012 L The work done by the gas during this process is: W = nRT ln(V2/V1) = (0.0018 kg)/(4.0 g/mol) * (8.314 J/mol*K) * ln(0.012 L/0.048 L) = -69 J Since ΔU = Q + W and ΔQ = 0 for an isothermal process, we have U3 = U2 - 69 J. Finally, for process 3 → 1, we can use the fact that the gas is undergoing an isobaric process (constant pressure), so we have: W = P(V1 - V3) = (1 atm)(0.048 L - 0.012 L) = 0.036 L*atm = 36 J Since ΔU = Q + W and ΔQ = 0 for an isobaric process, we have U1 = U3 + 36 J. Therefore, using the values we found, we can calculate the value of V₃: V₃ = U1 - U3 - W = (U3 + 36 J) - U3 - (-30 J) - (-69 J) = 135 J = 135 L*atm = 17 L*J. So the answer is A) 17 L.

The Figure (Not to Scale)shows a PV Diagram for 1