The gas inside of a container exerts $27 Pa$ of pressure and is at a temperature of $150 ^o K$ . If the pressure in the container changes to $40 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-11-20T06:40:07

The new temperature is $=222.2K$

Apply Gay Lussac's law

$P_1/T_1=P_2/T_2$ at constant volume

The initial pressure is $=P_1=27Pa$

The initial temperature is $=T_1=150K$

The final pressure is $P_2=40Pa$

The final temperature is

$T_2=P_2/P_1*T_1=40/27*150=222.2K$

Answer 2 · 2017-11-20T06:48:15

$220^@K$

Using the combined Gas Law, $(P_1V_1)/T_1=(P_2V_2)/T_2$ , you can simplify it to $P_1/T_1=P_2/T_2$ since volume is constant.

Since to want to find $T_2$ (Temperature 2), you rearrange the formula to become $T_2=(T_1*P_2)/P_1$

Then you can plug and chug
$T_2=(150^@K*40Pa)/(27Pa)$
$=222.222222222...^@K$
$=220^@K$ (with Significant Figures)

The gas inside of a container exerts 27 Pa27 Pa of pressure and is at a temperature of 150 ^o K150 ^o K. If the pressure in the container changes to 40 Pa40 Pa with no change in the container's volume, what is the new temperature of the gas?