We are given:
- volume of the tank (V1) = 15.7 L
- initial pressure in the tank (P1) = 237 atm
- final pressure in the tank (P2) = 1.00 atm
- volume of each balloon (V_balloon) = 2.00 L
- pressure in each balloon (P_balloon) = 1.00 atm
According to Boyle's law, the product of the initial pressure and volume of the gas is equal to the product of the final pressure and volume of the gas when the temperature remains constant. Therefore, (P1 * V1) = (P2 * V2).
Rearranging to find the final volume (V2), we get:
V2 = (P1 * V1) / P2
Now we can plug in the given values:
V2 = (237 atm * 15.7 L) / 1.00 atm
V2 = 3720.9 L
Now, we need to find the difference in volume between the initial and final states of the tank, which would be the volume available for inflation:
V_available = V2 - V1
V_available = 3720.9 L - 15.7 L
V_available = 3705.2 L
Finally, we need to calculate how many balloons can be inflated with this available volume. Since each balloon has a volume of 2.00 L, we can divide the available volume by the volume of one balloon:
No. of balloons = V_available / V_balloon
No. of balloons = 3705.2 L / 2.00 L
No. of balloons = 1852.6
Since we can't have a fraction of a balloon, we round down to the nearest whole number:
No. of balloons = 1852
A 15.7-L tank is filled with H2 to a pressure of 237 atm. How many balloons (each 2.00 L) can be inflated to a pressure of 1.00 atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below 1.00 atm pressure.
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