You’re at the zoo and have a big red 3.20 L helium balloon. The barometric pressure is 790.0 mm Hg. Then you hear the roar of a lion. Startled, you accidently release the balloon. It flies away. By the time it reaches the clouds the atmospheric pressure is only 0.300 atm. What is the new volume of the balloon?

The new volume, when the pressure drops to 0.300 atm, is 11.09 L

Explanation:

Step 1: Data given

The volume of the helium balloon = 3.20 L

The pressure in the balloon is 790.0 mmHg = 790 / 760 = 1.03947 atm

The pressure decreases to 0.300 atm

Step 2: Calculate the new volume

P1*V1 =P2*V2

⇒with P1 = the initial pressure = 1.03947 atm

⇒with V1 = the initial volume = 3.20 L

⇒with P2 = the reduced pressure = 0.300 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.03947 atm * 3.20 L = 0.300 atm * V2

V2 = (1.03947 atm * 3.20 L) / 0.300 atm

V2 = 11.09 L

The new volume, when the pressure drops to 0.300 atm, is 11.09 L

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point