Lögmál Boyle (fast hitastig) – vandamál og lausnir

1. Some ideal gases initially have þrýstingur P and volume V. If the gas undergoes isothermal process so that the final pressure becomes 4 times the initial pressure, then the final volume of gas is…

Þekkt:

Upphafsþrýstingur (P1) = P

Lokaþrýstingur (P2) = 4P

Initial volume (V1) = V

Óskað: Final volume (V2)

Lausn:

Formúlan af Lögmál Boyle :

P V = stöðug

P1 V1 = P2 V2

(P)(V) = (4P)(V2)

V = 4 V2

V2 = V / 4 = ¼ V

The final volume of gases is ¼ times the initial volume.

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2. In a closed container, the gas expands so that the endanleg rúmmál verður 2 times the initial volume (V = initial volume, P = initial pressure). The final pressure of gases is…

Þekkt:

Upphafsþrýstingur (P1) = P

Initial volume (V1) = V

Final volume (V2) = 2V

Óskast : Lokaþrýstingur (P2)

Lausn:

P1 V1 = P2 V2

P V = P2 (2V)

P= P2 (2)

P2 = P / 2 = ½ P

The gases pressure becomes ½ times the initial pressure.

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3. In a closed container, gases having a pressure of 2 atm and a volume of 1 liter. If gas pressure becomes 4 atm then gas volume becomes …

Þekkt:

Upphafsþrýstingur (P1) = 2 atm = 2 x 105 Pa

Lokaþrýstingur (P2) = 4 atm = 4 x 105 Pa

Initial volume (V1) = 1 liter = 1 dm3 = 1 x 10-3 m3

Óskast : Final volume (V2)

Lausn:

P1 V1 = P2 V2

(2 x 105)(1 x 10-3) = (4 x 105) V2

(1)(1 x 10-3) = (2) V2

1 x 10-3 = (2) V2

V2 = ½ x 10-3

V2 = 0.5 x 10-3 m3 = 0.5 dm3 = 0.5 lítris

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