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A child has a toy balloon with a volume of 2.90 liters. The temperature of the balloon when it was filled was 20.0 °C, and the pressure was 1.01 atm. If the child were to let go of the balloon, and it rose 3 kilometers into the sky where the pressure is 0.667 atm, and the temperature is -10.0°C, what would the new volume of the balloon be?

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Final answer:

Using the combined gas law and converting the temperatures to Kelvin, we can calculate the new volume of the balloon as approximately 3.99 liters when it rises 3 kilometers, where the pressure is 0.667 atm, and the temperature is -10.0°C.

Step-by-step explanation:

The question involves applying the combined gas law, which correlates pressure, volume, and temperature of a gas. Given that a child's toy balloon has a starting volume of 2.90 liters, at a temperature of 20.0 °C, and a pressure of 1.01 atm, we can calculate the new volume when the balloon is at a height where the pressure is 0.667 atm, and the temperature is -10.0°C.

To find the new volume, we use the combined gas law formula:
P1 × V1 / T1 = P2 × V2 / T2, where P is the pressure, V is the volume, and T is the temperature in Kelvin. Converting the temperatures to Kelvin, we get 293.15 K (20.0°C) and 263.15 K (-10.0°C) for T1 and T2, respectively. We re-arrange the formula to solve for V2:
V2 = P1 × V1 × T2 / P2 × T1. Plugging in the values:
V2 = (1.01 atm × 2.90 L × 263.15 K) / (0.667 atm × 293.15 K), which simplifies to V2 ≈ 3.99 liters.

Thus, the new volume of the balloon would be approximately 3.99 liters at the altitude of 3 kilometers.