Question

A latex balloon will adjust to maintain the same pressure. If the balloon is filled with 19.4 L of helium initially at 25.2 degrees Celcius, what would the volume be if the temperature adjusted to 5.4 degrees Celcius?

Answer 1

Answer:

V₂ =18.11  L

Explanation:

Given data:

Initial volume = 19.4 L

Initial temperature = 25.2°C (25.2+273 =298.2 K)

Final volume = ?

Final temperature = 5.4°C ( 5.4+273 =278.4K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 19.4 L × 278.4 K / 298.2 k

V₂ = 5400.96 L.K / 298.2 K

V₂ =18.11  L

Answer 2

hi, I'm lyo, a brazilian chemistry teacher ...

step by step :

The exercise can be solved using Charles' Law, which says: "that the amount of a gas in its volume provides the maximum number of moles in the pressure acquired"

The volume (mathematically speaking), is the amount of space occupies some type of body, in exercise it asks for the volume of the temperature adjusted to 5.4 ° C, so we have to:

V₁ / T₁ = V₂ / T₂ → the initial volume divided by the initial temperature, is equal to the final volume raised to (2), divided by the final temperature raised to (2) ...

Agora calculamos:

$V_{2} = \frac{ V_{1}T_{2}}{T_{1} } \\V_{2} = \frac{19,4 L * 278,4\; K }{298,2\; k} \\V_{2} = \frac{5400,96\; L\:K}{ 298,2\: K}\\ V_{2} = 18\: L$

the volume will then be 18 liters

$BRAINLY\; WE\; STUDY \;TOGETHER$