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A sample container of carbon monoxide occupies a volume of 435 ml at a pressure of 785 torr and a temperature of 298 k. What would its temperature be if the volume were changed to 265 ml at a pressure of 785 torr?

Respuesta :

opudodennis

Answer:

181.54 K

Explanation:

From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then

T2=v2t1/ v1

Given information

V1 435 ml

V2 265 ml

T1 298K

Substituting the given values then

T2=265*298/435=181.54 K

nuhulawal20

If the volume of the carbon monoxide was to change to the given volume, the temperature will also change to 181.5 Kelvins.

Given the data in the question;

  • Initial Volume; [tex]V_1 = 435ml[/tex]
  • Initial Temperature; [tex]T_1 = 298K[/tex]
  • Final Volume; [tex]V_2 = 265ml[/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{435ml}{298K} = \frac{265ml}{T_2}\\\\T_2 * 435ml = 298K * 265ml\\\\T_2 * 435ml = 78970ml.K\\\\T_2 = \frac{78970ml.K}{435ml } \\\\T_2 = 181.5K[/tex]

Therefore, if the volume of the carbon monoxide was to change to the given volume, the temperature will also change to 181.5 Kelvins.

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