At a birthday pool party, the temperature is 28.50°c and the atmospheric pressure is 755.4 mmhg. one of the decoration helium balloons has a volume of 6.25 l. what would be the volume of the balloon if it were submerged in a swimming pool to a depth where the pressure is 922.3 mmhg and the temperature is 26.35°c?

The volume of the balloon will be 5.11L

Explanation:

An excersise to solve with the Ideal Gases Law

First of all, let's convert the pressure in mmHg to atm

1 atm = 760 mmHg

760 mmHg ___ 1 atm

755.4 mmHg (755.4 / 760) = 0.993 atm

922.3 mmHg ( 922.3 / 760) = 1.214 atm

T° in K = 273 + °C

28.5 °C +273 = 301.5K

26.35°C + 273= 299.35K

P . V = n . R .T

First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K

(0.993atm . 6.25L) / 0.082 . 301.5 = n

0.251 moles = n

Second situation:

1.214 atm . V = 0.251 moles . 0.082 . 301.5K

V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm

V = 5.11L

The volume of the balloon if it were submerged in a swimming pool to a depth where the pressure is 922.3 mmHg and the temperature is 26.35°C is 5.08 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

where,

= initial pressure of gas = 755.4 mmHg

= final pressure of gas = 922.3 mmHg

= initial volume of gas = 6.25 L

= final volume of gas = ?

= initial temperature of gas =

= final temperature of gas =

Now put all the given values in the above equation, we get: