A perfectly spherical balloon occupied 35°C of neon gas under a pressure of 2 atm. The radius of the balloon is 15 cm. The balloon is then released up in the air until it
reached the temperature of -20°C under 0.3 atm. Determine its volume at that instant.

V₂ =  116126.75 cm³

Explanation:

Given data:

Radius of balloon = 15 cm

Initial pressure = 2 atm

Initial temperature = 35 °C (35 +273 = 308K)

Final temperature = -20°C (-20+273 = 253 K)

Final pressure = 0.3 atm

Final volume = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

Initial volume of balloon:

V = 4/3πr³

V = 4/3×22/7×(15cm)³

V = 14137.17 cm³

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2 atm × 14137.17 cm³ × 253 K / 308 K × 0.3 atm

V₂ = 7153408.02 atm .cm³. K / 61.6 K.atm

V₂ =  116126.75 cm³

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.