A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? Assume that the temperature remains constant
To find the volume of the helium gas when the pressure changes, we can use the combined gas law equation, which states:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
- P1 and P2 are the initial and final pressures, respectively,
- V1 and V2 are the initial and final volumes, respectively,
- T1 and T2 are the initial and final temperatures, respectively.
Given:
- P1 = 103 kPa (initial pressure)
- V1 = 30.0 L (initial volume)
- P2 = 25.0 kPa (final pressure)
- T1 = T2 (temperature remains constant)
We need to solve for V2 (final volume).
Rearranging the equation, we have:
V2 = (P1 * V1 * T2) / (P2 * T1)
Since the temperature remains constant, T1 = T2 and we can simplify the equation to:
V2 = (P1 * V1) / P2
Now, substitute the given values:
V2 = (103 kPa * 30.0 L) / 25.0 kPa
Calculating this equation, we find:
V2 ≈ 123.6 L
Therefore, the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa is approximately 123.6 L.