A weather balloon has a volume of 200.0 L at a pressure of 760 mm Hg. As it rises, the pressure decreases to 282 mm Hg. a. Will the volume of the balloon increase or decrease? Why?

b. What is the new volume of the balloon? (Assume constant temperature)

a. Increase

b. 539 L

Explanation:

Step 1: Given data

Step 2: Calculate the final volume of the gas

The pressure of the gas is inversely proportional to the volume. So, as the pressure decreases, the volume must increase. Assuming constant temperature and ideal behavior, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 760 mmHg × 200.0 L/282 mmHg = 539 L

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2