The gas inside of a container exerts $12 P a$ $12 Pa$ of pressure and is at a temperature of $120^{o} K$ $120 ^o K$ . If the temperature of the gas changes to $60^{o} K$ $60 ^oK$ with no change in the container's volume, what is the new pressure of the gas?

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Answer 1 · 2018-01-21T18:05:56

The new pressure is $= 6 P a$ $=6Pa$

Apply Gay Lussac 's Law

$\frac{Pressure}{Temperature} = Constant$ $"Pressure"/"Temperature"="Constant"$ at $Constant Volume$ $"Constant Volume"$

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1=P_2/T_2$

The initial pressure is $P_{1} = 12 P a$ $P_1=12Pa$

The initial temperature is $T_{1} = 120 K$ $T_1=120K$

The final temperature is $T_{2} = 60 K$ $T_2=60K$

The final pressure is

$P_{2} = \frac{T_{2}}{T_{1}} \cdot T_{2}$ $P_2=T_2/T_1*T_2$

$= \frac{60}{120} \cdot 12$ $=60/120*12$

$= 6 P a$ $=6Pa$

The gas inside of a container exerts 12Pa12 Pa of pressure and is at a temperature of 120oK120 ^o K. If the temperature of the gas changes to 60oK60 ^oK with no change in the container's volume, what is the new pressure of the gas?