A weather balloon is inflated to a volume of 29.6 L at a pressure of 734 mmHg and a temperature of 27.0 C. The balloon rises in the atmosphere to an altitude, where the pressure is 360 mmHg and the temperature is -13.6 C.
To solve this problem, we can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
We are given the initial volume (V1 = 29.6 L), initial pressure (P1 = 734 mmHg), and initial temperature (T1 = 27.0 C).
Using the ideal gas law, we can calculate the number of moles (n1) of air inside the balloon at the initial conditions. We need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature:
T1 (Kelvin) = 27.0 + 273.15 = 300.15 K
Now, let's calculate n1. Rearranging the ideal gas law equation gives us:
n1 = (P1 * V1) / (R * T1)
The ideal gas constant, R, is 0.0821 L•atm/(mol•K) or 62.36 mmHg•L/(mol•K). Since the pressure is given in mmHg, we'll use the second value for R.
n1 = (734 mmHg * 29.6 L) / (62.36 mmHg•L/(mol•K) * 300.15 K)
Simplifying the equation gives us:
n1 ≈ 0.377 moles of air
Now, we need to find the final volume of the balloon (V2), given the final pressure (P2 = 360 mmHg) and final temperature (T2 = -13.6 C).
Just like before, we need to convert T2 to Kelvin:
T2 (Kelvin) = -13.6 + 273.15 = 259.55 K
Using the ideal gas law equation again, we can calculate the final volume of the balloon (V2):
V2 = (n2 * R * T2) / P2
Since the number of moles (n2) remains constant, we can rearrange the equation to solve for V2:
V2 = (n1 * R * T2) / P2
Substituting all the known values:
V2 ≈ (0.377 mol * 62.36 mmHg•L/(mol•K) * 259.55 K) / 360 mmHg
Simplifying the equation gives us:
V2 ≈ 0.618 L
Therefore, the final volume of the balloon at the given altitude is approximately 0.618 L.