The gas inside of a container exerts $15 Pa$ of pressure and is at a temperature of $450 ^o K$ . If the pressure in the container changes to $32 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-03-14T03:42:46

$T_2= 960 K$

Lussac's Law:
$P_1/T_1=P_2/T_2$

$T_2=(P_2*T_1)/P_1$

$T_2=(32Pa*450K)/(15 Pa)$

$T_2= 960 K$

The gas inside of a container exerts 15 Pa15 Pa of pressure and is at a temperature of 450 ^o K450 ^o K. If the pressure in the container changes to 32 Pa32 Pa with no change in the container's volume, what is the new temperature of the gas?