The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #60 ^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

The temperature is #=106.7K#

We use Gay Lussac's Law

#P_1/T_1=P_2/T_2#
#P_1=18Pa#
#T_1=60K#
#P_2=32Pa#
#T_2=(P_2T_1)/P_1#
#=32*60/18=106.7K#
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Answer 2

The new temperature of the gas is 107.33 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.

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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