The gas inside of a container exerts $18 Pa$ of pressure and is at a temperature of $70 ^o K$ . If the pressure in the container changes to $48 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-02-19T10:57:31

The new trmperature is $=186.7K$

We apply Gay Lussac's Law

$P_1/T_1=P_2/T_2$

$P_1=18Pa$

$T_1=70K$

$P_2=48Pa$

$T_2=P_2/P_1*T_1$

$=48/18*70=186.7K$

The gas inside of a container exerts 18 Pa18 Pa of pressure and is at a temperature of 70 ^o K70 ^o K. If the pressure in the container changes to 48 Pa48 Pa with no change in the container's volume, what is the new temperature of the gas?