The gas inside of a container exerts #25 Pa# of pressure and is at a temperature of #450 ^o K#. If the temperature of the gas changes to #550 ^oK# with no change in the container's volume, what is the new pressure of the gas?

Answer 1

#P~=30,56 Pa#

#25/450=P/550# #P=(25*550)/450# #P~=30,56 Pa#
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Answer 2

Use the Gay-Lussac's Law formula to find the new pressure ((P_2)) when the temperature changes:

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

Given (P_1 = 25 , \text{Pa}), (T_1 = 450 , \text{K}), and (T_2 = 550 , \text{K}), solve for (P_2):

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

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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.

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.

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.

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.

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