Respuesta :
V₁ = 400.0 mL
T₁ = 22.0ºC + 273 = 295 K
V₂ = ?
T₂ = 30.0ºC + 273 = 303 K
V₁ / T₁ = V₂ / T₂
400.0 / 295 = V₂ / 303
295 x V₂ = 400.0 x 303
295 V₂ = 121200
V₂ = 121200 / 295
V₂ = 410.84 mL
hope this helps!
T₁ = 22.0ºC + 273 = 295 K
V₂ = ?
T₂ = 30.0ºC + 273 = 303 K
V₁ / T₁ = V₂ / T₂
400.0 / 295 = V₂ / 303
295 x V₂ = 400.0 x 303
295 V₂ = 121200
V₂ = 121200 / 295
V₂ = 410.84 mL
hope this helps!
Considering the Charles' Law, the volume increases from 400 mL to 410.85 mL as the temperature increases from 22.0 °C to 30.0°C.
Charles' Law
Charles' Law consists of the relationship that exists between the volume and the temperature of a certain amount of ideal gas, which is maintained at a constant pressure.
For a given sum of gas, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. This is, the volume is directly proportional to the temperature of the gas:
In summary, Charles' law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:
[tex]\frac{V}{T} = k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is fulfilled:
[tex]\frac{V1}{T1} = \frac{V2}{T2}[/tex]
Volume change
In this case, you know:
- V1= 400 mL
- T1= 22 C= 295 K (being 0 C= 273 K)
- V2= ?
- T2= 30 C= 303 K
Replacing in the Charles's law:
[tex]\frac{400 mL}{295 K} = \frac{V2}{303 K}[/tex]
Solving:
[tex]\frac{400 mL}{295 K}x303 K = V2[/tex]
410.85 mL= V2
Finally, the volume increases from 400 mL to 410.85 mL as the temperature increases from 22.0 °C to 30.0°C.
Learn more about Charles's law:
https://brainly.com/question/4147359