The gas inside of a container exerts $6 Pa$ of pressure and is at a temperature of $120 ^o K$ . If the pressure in the container changes to $49 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-03-21T14:42:52

The new temperature is $=980K$

Apply Gay Lussac's Law

$"Pressure"/"Temperature"= "Constant"$ , at $"Constant Volume"$

$P_1/T_1=P_2/T_2$

The initial pressure is $P_1=6Pa$

The initial temperature is $T_1=120K$

The final pressure is $P_2=49Pa$

The final temperature is

$T_2=P_2/P_1*T_1$

$=49/6*120=980K$

The gas inside of a container exerts 6 Pa6 Pa of pressure and is at a temperature of 120 ^o K120 ^o K. If the pressure in the container changes to 49 Pa49 Pa with no change in the container's volume, what is the new temperature of the gas?