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The gas inside of a container exerts #3 Pa# of pressure and is at a temperature of #430 ^o K#. If the temperature of the gas changes to #120 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Answer 1

Charlotte Aaron

#P~=2,742 Pa#

#"use Gay - Lussac law"# #120 ^o C=120+273=393 ^o K# #3/430=P/393# #P=(3*393)/430# #P~=2,742 Pa#

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Answer 2

Parker Abbott

To find the new pressure of the gas when the temperature changes, we use the ideal gas law equation:

[ P_1 \times V_1 = P_2 \times V_2 ]

Given:

Initial pressure, ( P_1 = 3 , \text{Pa} )
Initial temperature, ( T_1 = 430 , \text{K} )
Final temperature, ( T_2 = 120 ^\circ C = 120 + 273 = 393 , \text{K} )
Volume remains constant, so ( V_1 = V_2 )

We rearrange the equation to solve for ( P_2 ):

[ P_2 = \frac{P_1 \times T_2}{T_1} ]

[ P_2 = \frac{3 , \text{Pa} \times 393 , \text{K}}{430 , \text{K}} ]

[ P_2 \approx 2.748 , \text{Pa} ]

Therefore, the new pressure of the gas is approximately ( 2.748 , \text{Pa} ).

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Answer from HIX Tutor

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