To solve this problem, we can use the combined gas law, which combines Boyle's law, Charles's law, and Gay-Lussac's law. The combined gas law formula is:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 = Initial pressure (in mm Hg)
V1 = Initial volume (in mL)
T1 = Initial temperature (in Kelvin)
P2 = Final pressure (in mm Hg)
V2 = Final volume (in mL)
T2 = Final temperature (in Kelvin)
Given data:
P1 = 322.5 mm Hg
V1 = 280 mL
T1 = 26 °C = 26 + 273.15 = 299.15 K (converted to Kelvin)
V2 = 716.4 mL
T2 = 69.1 °C = 69.1 + 273.15 = 342.25 K (converted to Kelvin)
Now, let's plug in the values and solve for P2:
(P1 * V1) / T1 = (P2 * V2) / T2
(322.5 * 280) / 299.15 = (P2 * 716.4) / 342.25
Solve for P2:
P2 = (322.5 * 280 * 342.25) / (299.15 * 716.4)
P2 ≈ 144 mm Hg
Therefore, the pressure exerted by the gas at 69.1 °C and a volume of 716.4 mL is approximately 144 mm Hg. The correct answer is option "a. 144".