The gas inside of a container exerts $12 P a$ $12 Pa$ of pressure and is at a temperature of $85^{o} K$ $85 ^o K$ . 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 · 2017-11-27T07:06:04

The new pressure is $= 8.47 P a$ $=8.47Pa$

Apply Gay Lussac's Law, which is

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1=P_2/T_2$ , at constant volume

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

The initial temperature is $T_{1} = 85 K$ $T_1=85K$

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

The final pressure is

$P_{2} = \frac{T_{2}}{T_{1}} \cdot P_{1} = \frac{60}{85} \cdot 12 = 8.47 P a$ $P_2=T_2/T_1*P_1=60/85*12=8.47Pa$

The gas inside of a container exerts 12 Pa12Pa12 Pa of pressure and is at a temperature of 85 ^o K85oK85 ^o K. If the temperature of the gas changes to 60 ^oK60oK60 ^oK with no change in the container's volume, what is the new pressure of the gas?