Charles’s law (constant pressure) – problems and solutions

1. In a closed container, the gas expands so that the final volume becomes 3 times the initial volume (V = initial volume, T = initial pāmahana). What is the final temperature?

Mōhiotia:

Te rōrahi tīmatanga (V1) = V

Rōrahi whakamutunga (V2) = 3V

Initial temperature (T1) = T

Hiahia: Final temperature (T2)

Rongoā:

Ko te tātai o Charles’s law :

Charles's law (constant pressure) - problems and solutions 1

The final temperature of gases becomes 3 times the initial temperature.

A hi'o atoa  Ngā ara iahiko hiko – ngā raruraru me ngā otinga

2. Ahaudeal gases initially have volume V and temperature T. If the gas undergoes the isobaric process so that the temperature becomes 2 times the initial temperature then the final volume of gases is…

Mōhiotia:

Te rōrahi tīmatanga (V1) = V

Initial temperature (T1) = T

Final temperature (T2) = 2T

Hiahia: final volume (V2)

Rongoā:

Charles's law (constant pressure) - problems and solutions 2

The final volume of gases becomes 2 times the initial volume.

A hi'o atoa  Ngā pūnga iahiko raupapa me te whakarara – ngā raruraru me ngā otinga

3. In a closed container, ideal gases initially have a volume of 2 liters and temperature of 27oC. If the final volume of gases becomes 3 liters then the final temperature is…

Mōhiotia:

Te rōrahi tīmatanga (V1) = 2 liters = 2 dm3 = 2x10-3 m3

Rōrahi whakamutunga (V2) = 3 liters = 3 dm3 = 3x10-3 m3

Initial temperature (T1) = 27oC + 273 = 300 K

E hiahiatia ana: Final temperature (T2)

Rongoā:

Charles's law (constant pressure) - problems and solutions 3

The final temperature is 177oC or 177 + 273 = 450 Kelvin.

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