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A gas sample is held at constant pressure. The gas occupies 3.62 L of volume when the temperature is 21.6°C. Determine the temperature at which the volume of the gas is 3.45 L.

Respuesta :

Niviinsoul

Answer:

281 K

Explanation:

Charles's Law. V1/T1 = V2/T2.

The temperature must be in K = 21.6°C + 273 = 294.6K.

V1T2 = V2T1.

3.62L x T2 = 3.45L x 294.6K

T2 = (3.45 x 294.6) / 3.62 = 1016.4 / 3.62 = B): 281K.

(By direct proportion of volume change: (3.45L / 3.62L) x 294.6K = 281K).

nuhulawal20

The temperature at which the volume of the gas is 3.45L is 280.908K or 7.76°C.

Given the data in the question;

  • Initial Volume; [tex]V_1 = 3.62L[/tex]
  • Initial Temperature; [tex]T_1 = 21.6^oC = ( 21.6 + 273.15)K = 294.75K[/tex]
  • Final Volume; [tex]V_2= 3.45L[/tex]

Initial Temperature; [tex]T_2 =\ ?[/tex]

To determine the final temperature, we use the Equation from Charles's Law:

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

We substitute our given values into the equation

[tex]\frac{3.62L}{294.75K} = \frac{3.45L}{T_2}\\\\T_2 * 3.62L = 294.75K * 3.45L\\\\T_2 * 3.62L = 1016.8875L.K\\\\T_2 = \frac{1016.8875L.K}{3.62L} \\\\T_2 = 280.908K\\\\T_2 = ( 280.908 - 273.15) ^oC = 7.76^oC[/tex]

Therefore, the temperature at which the volume of the gas is 3.45L is 280.908K or 7.76°C.

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