A helium balloon is filled to a volume of 2.88 x 103 L at 722 mm Hg and 19°C. When the balloon is released, it rises to an altitude where the pressure and temperature are 339 mm Hg and -55oC, respectively. What is the volume of the balloon (in L)?
Since PV/T is constant, you want V such that
(339*V) / (273-55) = (722 * 2.88*10^3) / (273+19)
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of pressure, temperature, and volume. The combined gas law formula is given as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Given values:
P1 = 722 mm Hg
V1 = 2.88 x 10^3 L
T1 = 19°C (which is equivalent to 19 + 273 = 292 K)
P2 = 339 mm Hg
T2 = -55°C (which is equivalent to -55 + 273 = 218 K)
Let’s substitute these values into the equation:
(722 mm Hg * 2.88 x 10^3 L) / (292 K) = (339 mm Hg * V2) / (218 K)
Rearranging the equation to solve for V2:
V2 = (722 mm Hg * 2.88 x 10^3 L * 218 K) / (339 mm Hg * 292 K)
Now we can calculate the volume:
V2 = (722 * 2.88 x 10^3 * 218) / (339 * 292)
V2 ≈ 1654.08 L
Therefore, the volume of the balloon, when it reaches the new altitude, is approximately 1654.08 L.
To find the volume of the balloon at the new conditions, we can use the combined gas law equation:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume (unknown)
T2 = final temperature
Let's plug in the values we have:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
(722 mm Hg × 2.88 x 10³ L) / (19°C + 273) = (339 mm Hg × V2) / (-55°C + 273)
Now, we can solve this equation for V2 by cross-multiplication:
(722 mm Hg × 2.88 x 10³ L) × (273 - 55°C) = (339 mm Hg) × (19°C + 273) × V2
V2 = [(722 mm Hg × 2.88 x 10³ L) × (273 - 55°C)] / [(339 mm Hg) × (19°C + 273)]
Now, calculate the value of V2 using the given values:
V2 = [(722 mm Hg × 2.88 x 10³ L) × (273 - 55°C)] / [(339 mm Hg) × (19°C + 273)]
V2 ≈ 2.31 x 10³ L
Therefore, the volume of the balloon at the new altitude is approximately 2.31 x 10³ L.