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A gas expands from a volume of 2.00L at 36.0oC to a volume of 2.50 L, what is the final temperature, if the pressure is constant?

Respuesta :

teamath333
We shall consider V, the volume and T, the temperature.
According to Boyle's Laws:
[tex] \frac{V1}{T1} = \frac{V2}{T2} [/tex]
In our case:
[tex] \frac{2.00}{36.0} = \frac{2.50}{T2} =\ \textgreater \ T2= \frac{2.50*36.00}{2.00} = 45 [/tex]
nandhini123

The final temperature is 45.45 Celcius

Explanation:

The Combined Gas Law:

The combined gas law allows to derive relationships between  the variable that undergoes like pressure, temperature and volume.

[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]

It is given thatpressure is constant so,

[tex]P_1=P_2[/tex]

Hence combined gas law becomes,

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

Substituting the values given in the question,

[tex]\frac{2.00}{36}=\frac{2.50}{T_2}[/tex]

[tex]0.055=\frac{2.50}{T_2}[/tex]

[tex]T_2=\frac{2.50}{0.055}[/tex]

[tex]T_2[/tex]=45.45 C

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