Um gás ideal inicialmente a 0,85 atm e 66 ·C, expande-se até que seu volume, pressão e temperatura sejam respectivamente 94 mL, 0.60 atm e 45 °C_ Sob tais condições, o volume inicial desse gás era de, aproximadamente:

Since this question was annulled and does not have a correct answer, I will explain the reasoning behind it and how to approach this type of problem.

The given information states that the ideal gas initially has a pressure of 0.85 atm and a temperature of 66°C. After expansion, the volume, pressure, and temperature become 94 mL, 0.60 atm, and 45°C, respectively. To find the initial volume, we can use the ideal gas law:

$$PV = nRT$$

Since the number of moles (n) and the gas constant (R) are not provided, we can assume that they are constant. Rearranging the equation to solve for V, we get:

$$V = frac{nRT}{P}$$

Using the initial conditions, we can write:

$$V_{initial} = frac{nRT_{initial}}{P_{initial}}$$

And using the final conditions, we can write:

$$V_{final} = frac{nRT_{final}}{P_{final}}$$

Since the number of moles (n) and the gas constant (R) are constant, we can equate the two expressions:

$$frac{nRT_{initial}}{P_{initial}} = frac{nRT_{final}}{P_{final}}$$

Simplifying the equation, we get:

$$frac{T_{initial}}{P_{initial}} = frac{T_{final}}{P_{final}}$$

Plugging in the given values, we get:

$$frac{66°C + 273.15}{0.85 atm} = frac{45°C + 273.15}{0.60 atm}$$

Solving for the initial volume, we get:

$$V_{initial} = 50.4 mL$$

However, since the question was annulled, this answer is not considered correct. Nevertheless, this is how you would approach this type of problem using the ideal gas law and the given conditions.