The gas inside of a container exerts #15 Pa# of pressure and is at a temperature of #450 ^o K#. If the pressure in the container changes to #32 Pa# with no change in the container's volume, what is the new temperature of the gas?

Answer 1

Charles Aeschliman

#T_2= 960 K#

Lussac's Law: #P_1/T_1=P_2/T_2#

#T_2=(P_2*T_1)/P_1#

#T_2=(32Pa*450K)/(15 Pa)#

#T_2= 960 K#

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Answer 2

Adrian Ayala

Using Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature when the volume is constant, we can set up the following equation:

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

Given ( P_1 = 15 , \text{Pa} ), ( T_1 = 450 , \text{K} ), and ( P_2 = 32 , \text{Pa} ), we can solve for ( T_2 ):

[ \frac{15}{450} = \frac{32}{T_2} ]

[ T_2 = \frac{32 \times 450}{15} = 960 , \text{K} ]

Therefore, the new temperature of the gas is ( 960 , \text{K} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.