What pressure would it take to compress 300. L Of Helium at 1.00atm into a 2.00 L tank at constant temperature

150 atm

Explanation:

Given data:

Initial volume = 300 L

Initial pressure = 1 atm

Final volume = 2.00 L

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

1 atm × 300 L = P₂ × 2.00 L

P₂ = 300 atm. L/ 2.00 L

P₂ = 150 atm

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3