The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #690 ^o K#. If the temperature of the gas changes to #210 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Answer 1

The new pressure is #=6.3Pa#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#
#P_1=9Pa#
#T_1=690K#
#T_2=210+273=483K#

So,

#P_2=T_2/T_1*P_1#
#=483/690*9#
#=6.3Pa#
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Answer 2

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

[P_1/T_1 = P_2/T_2]

Given: [P_1 = 9 , \text{Pa}] (initial pressure) [T_1 = 690 , \text{K}] (initial temperature) [T_2 = 210^\circ C = 210 + 273 = 483 , \text{K}] (new temperature)

Let's plug these values into the equation and solve for (P_2):

[9 , \text{Pa} / 690 , \text{K} = P_2 / 483 , \text{K}]

[P_2 = (9 , \text{Pa} \times 483 , \text{K}) / 690 , \text{K}]

[P_2 = 6.32 , \text{Pa}]

Therefore, the new pressure of the gas is (6.32 , \text{Pa}) when the temperature changes to (210^\circ C).

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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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