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A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.

If the gas was cooled from 20°C to -60°C, what was its new volume? ( Don't forget to convert Celsius to Kelvin)

Respuesta :

joseaaronlara

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               [tex]\frac{V1}{T1} = \frac{V2}{T2}[/tex]

Solve for V2

                V2 = [tex]\frac{V1T2}{T1}[/tex]

3.- Substitution

               V2 = [tex]\frac{(42)(233)}{293}[/tex]

4.- Simplification

                V2 = [tex]\frac{9786}{293}[/tex]

5.- Result

                V2 = 33.4ml

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