Respuesta :
Answer:
Therefore,
The final kelvin temperature when the volume of the gas is changed is
[tex]T_{2}=566\ K[/tex]
Explanation:
Given:
[tex]V_{1}=444\ ml\\T_{1}=273\ K\\P_{1}=79\ kPa[/tex]
[tex]V_{2}=1880\ ml\\P_{2}=38.7\ kPa[/tex]
To Find:
[tex]T_{2}=?[/tex]
Solution:
Combined Gas Law:
The combined gas law combines the three gas laws:
Boyle's Law, Charles' Law, and Gay-Lussac's Law.
It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
Hence,
[tex]\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}[/tex]
Substituting the values we get
[tex]T_{2}=\dfrac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}[/tex]
[tex]T_{2}=\dfrac{38.7\times 1880\times 273}{79\times 444}=566.26\approx 566\ K[/tex]
Therefore,
The final kelvin temperature when the volume of the gas is changed is
[tex]T_{2}=566\ K[/tex]
