Given constant pressure, how does the temperature change when a $500cm^3$ volume of gas at $373K$ is cooled so that then volume reduces to $270*cm^3$ ?

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Answer 1 · 2017-08-28T13:59:48

Old Charles' Law holds that $V_1/T_1=V_2/T_2$ under conditions of constant pressure......

We solve for $T_2=V_2/V_1xxT_1=(270*cm^3)/(500*cm^3)xx373*K$

$=201.5*K$

Clearly the temperature has to drop to reduce the volume so substantially.

Given constant pressure, how does the temperature change when a 500*cm^3500*cm^3 volume of gas at 373*K373*K is cooled so that then volume reduces to 270*cm^3270*cm^3?