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The gas inside of a container exerts #4 Pa# of pressure and is at a temperature of #120 ^o C#. If the temperature of the gas changes to #150 ^oK# with no change in the container's volume, what is the new pressure of the gas?

Answer 1

Adrian Ayala

The new pressure is #=1.53Pa#

We utilize the Gay Lussac Law.

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=4Pa#

The initial temperature is #T_1=120+273=393K#

The final temperature is #T_2=150K#

The last bit of pressure is

#P_2=T_2/T_1*P_1#

#=150/393*4#

#=1.53Pa#

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Answer 2

Amelia Adams

To find the new pressure of the gas, we can use the ideal gas law:

[ P_1/T_1 = P_2/T_2 ]

Where: ( P_1 = 4 , \text{Pa} ) (initial pressure) ( T_1 = 120^\circ \text{C} = 393 , \text{K} ) (initial temperature) ( T_2 = 150^\circ \text{C} = 423 , \text{K} ) (final temperature)

Plugging in the values:

[ 4/393 = P_2/423 ]

[ P_2 = (4 \times 423)/393 ]

[ P_2 ≈ 4.31 , \text{Pa} ]

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Answer from HIX Tutor

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