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What is the volume of gas at 2.00 atm and 200.0 K if it’s original volume was 300.0 L at 0.250 atm and 400.0 K.

Respuesta :

Answer:

= 18.75 L

Explanation:

Using the combined gas law;

P1V1/T2=P2V2/T2

Where; P1 is the initial pressure, 0.25 atm

V1 is the initial volume 300.0 L

T1 is the initial temperature, 400 K

P2 is 2 atm and

T2 is 200 K

Therefore;

(0.25 × 300)/400 = 2V2/200

V2 = (75 ×200)/(400×2)

     = 18.75 L

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