contestada

Calcium Carbonate decomposes at 1200°C to form carbon dioxide and
calcium oxide. If 25 L of carbon dioxide are collected at 1200°C, what will
the volume of this gas be after it cools to 25°C?

Respuesta :

This is a Charles' Law problem: V1/T1 = V2/T2. As the temperature of a fixed mass of gas decreases at a constant pressure, the volume of the gas should also decrease proportionally.

To use Charles' Law, the temperature must be in Kelvin (x °C = x + 273.15 K). We want to solve Charles' Law for V2, which we can obtain by rearranging the equation into V2 = V1T2/T1. Given V1 = 25 L, T1 = 1200 °C (1473.15 K), and T2 = 25 °C (298.15 K):

V2 = (25 L)(298.15 K)/(1473.15 K) = 5.1 L.  

Cricetus

The volume of this gas will be "5.1 L".

Given:

  • [tex]V_1 = 25 \ L[/tex]
  • [tex]T_1 = (1200+ 273 )K[/tex]

             [tex]= 1473 \ K[/tex]

  • [tex]T_2 = (25+ 273)K[/tex]

             [tex]= 298 \ K[/tex]

As we know,

→ [tex]\frac{V_1}{V_2} = \frac{T_1}{T_2}[/tex]

By substituting the values, we get

→ [tex]\frac{25 \ L}{V_2} = \frac{1473 \ K}{293 \ K}[/tex]

→    [tex]V_2 = \frac{25\times 298}{1473} \ L[/tex]

→         [tex]= 5.1 \ L[/tex]

Learn more:

https://brainly.com/question/21437107

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