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A person has a sample of gas with a volume of 7.41L and a temperature of 418.55K. If the volume of the container is reduced to 2.59L, what will the new temperature of the gas be in Kelvin?

Respuesta :

For this problem, we could use the Charles's law:

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

We could replace the values that the question gives to us:

[tex]\frac{7.41L}{418.55K}=\frac{2.59L}{T_2}\to T_2=\frac{418.55K\cdot2.59L}{7.41L}=146.29K[/tex]

Therefore, the new temperature of the gas is 146.29K.

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