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A gas that has a volume of 28 L, a temperature of 45 °C, and an unknown pressure has its volume increased to 34 L and its temperature decreased to 35 °C. If the pressure is measured after the change to be 2.0 atm, what was the original pressure of the gas?A.1.6 atmB.2.5 atmC.3.2 atmD.4.1 atm

Respuesta :

joseaaronlara

Answer:

The answer to your question is letter B. 2.5 atm

Explanation:

Data

P1 = ?                  P2 = 2 atm

T1 = 45°C = 318°K           T2 = 35°C = 308°K

V1 = 28 L             V2 = 34 L

Formula

Combined gas law

   P1V1/T1 = P2V2/T2

solve for P1

   [tex]P1 = \frac{T1P2V2}{V1T2}[/tex]

Substitution

  P1 = (318 x 2 x 34) / (28 x 308)

Simplification

  P1 = 21624 / 8624

Result

  P1 = 2.5 atm

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