Question

A helium-filled weather balloon has a volume of 512 L at 18.9°C and 756 mmHg. It is released and rises to an altitude of 2.14 km, where the pressure is 639 mmHg and the temperature is 5.9°C.
The volume of the balloon at this altitude is
L.

Answer 1

Answer:

578.8 L

Explanation:

In order to solve this problem, we can use the equation of state for an ideal gas, which states that:

$pV=nRT$

where

p is the pressure of the gas

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

The equation can also be rewritten as

$\frac{pV}{T}=nR$

For a gas transformation, the term on the right (nR) is constant, so we can write:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where in this problem:

$p_1 = 756 mm Hg$ is the initial pressure of the gas

$V_1=512 L$ is the initial volume

$T_1=18.9^{\circ}+273=291.9 K$ is the initial temperature

$p_2 = 639 mm Hg$ is the final pressure

$T_2=5.9^{\circ}C+273=278.9 K$ is the final temperature

Solving for V2, we find the final volume of the balloon:

$V_2=\frac{p_1 V_1T_2}{T_1p_2}=\frac{(756)(512)(278.9)}{(291.9)(639)}=578.8 L$

Answer 2

633.97 L

Explanation:

Well use the combined gas law;

P₁V₁T₁ = P₂V₂T₂

We need to change the temperatures into Kelvin;

18.9°C= 292.05 K

5.9°C = 279.05 K

756 * 512 * 292.05 = 639 * V₂ * 279.05

113,044,377.6 = 178,312.95 V₂

V₂ = 113,044,377.6 / 178,312.95

V₂ = 633.97 L