Calculate the temperature of gas that originally occupied 3.45L and is expanded to 5.25L. The original temperature of the gas was 282K and the pressure remains constant.

Respuesta :

V1/V2=T1/T2
[tex] \frac{V1}{ \:V2} = \frac{T1}{T2} \\ \frac{3.45}{5.25} = \frac{282}{T2} \\ T2 = \frac{5.25 \times 282}{3.45} \\ T2 = 429 \: k[/tex]

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