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)?

Question

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)?

Answer 1

You want V such that

339V/(273-55) = (722 * 2.88*10^3)/(273+19)

Answer 2

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 of the balloon (722 mm Hg)
V1 = initial volume of the balloon (2.88 x 10^3 L)
T1 = initial temperature of the balloon (19°C + 273.15 = 292.15 K)
P2 = final pressure of the balloon (339 mm Hg)
V2 = final volume of the balloon (to be calculated)
T2 = final temperature of the balloon (-55°C + 273.15 = 218.15 K)

Now, we can plug in the values into the equation and solve for V2:

(722 mm Hg * 2.88 x 10^3 L) / 292.15 K = (339 mm Hg * V2) / 218.15 K

(722 * 2.88 x 10^3) / 292.15 = (339 * V2) / 218.15

2081280 / 292.15 = (339 * V2) / 218.15

7112.0221 = (339 * V2) / 218.15

Multiplying both sides by 218.15:

7112.0221 * 218.15 = 339 * V2

1552422.81315 = 339 * V2

Dividing both sides by 339:

V2 = 1552422.81315 / 339

V2 ≈ 4580.90 L (rounded to two decimal places)

Therefore, the volume of the balloon at the new conditions is approximately 4580.90 L.

Answer 3

To solve this question, we can use the combined gas law equation:

(P₁V₁) / (T₁) = (P₂V₂) / (T₂)

Let's assign the given values to the variables:
P₁ = 722 mm Hg (initial pressure)
V₁ = 2.88 x 10³ L (initial volume)
T₁ = 19°C + 273.15 = 292.15 K (initial temperature)

P₂ = 339 mm Hg (final pressure)
V₂ = ? (final volume, what we need to find)
T₂ = -55°C + 273.15 = 218.15 K (final temperature)

Now we can substitute these values into the combined gas law equation:

(722 mm Hg * 2.88 x 10³ L) / (292.15 K) = (339 mm Hg * V₂) / (218.15 K)

Let's simplify the equation:

(2090496 mm Hg * L) / (292.15 K) = (339 mm Hg * V₂) / (218.15 K)

Now we can solve for V₂ by cross multiplying:

(2090496 * 218.15 K) / (292.15 K) = 339 mm Hg * V₂

Simplifying further:

V₂ = [(2090496 * 218.15 K) / (292.15 K)] / 339 mm Hg

Using a calculator, we find:

V₂ ≈ 1550.516 L

Therefore, the volume of the balloon at the new altitude is approximately 1550.516 L.