Question

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?

Answer 1

Answer: 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

Step-by-step explanation:

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

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 755.4 mmHg

`P_2` = final pressure of gas = 922.3 mmHg

`V_1` = initial volume of gas = 6.25 L

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`28.50^oC=273+28.50=301.50K`

`T_2` = final temperature of gas =
`26.35^oC=273+26.35=299.35K`

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

`(755.4 * 6.25)/(301.50K)=(922.3* V_2)/(299.35K)`

`V_2=5.08L`

Answer 2

The volume of the balloon will be 5.11L

Step-by-step 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