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DanielleElmas [232]

3 years ago

5

At a birthday pool party, the temperature is 28.50°C and the atmospheric pressure is 755.4 mmHg. One of the decoration helium ba

lloons 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?

2 answers:

Iteru [2.4K]3 years ago

7 0

Answer:

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

vagabundo [1.1K]3 years ago

6 0

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

Explanation:

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

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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:

$\frac{755.4 \times 6.25}{301.50K}=\frac{922.3\times V_2}{299.35K}$

$V_2=5.08L$

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