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The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #190 ^o K#. If the pressure in the container changes to #34 Pa# with no change in the container's volume, what is the new temperature of the gas?

Answer 1

Benjamin Asiedu

The temperature is #=717.8K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=9Pa#

#T_1=190K#

#P_2=34Pa#

#T_2=(P_2*T_1)/P_1#

#=34*190/9=717.8K#

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Answer 2

Nathan Ashcroft

Use the combined gas law: ( P_1/T_1 = P_2/T_2 ) Rearrange the equation to solve for ( T_2 ): ( T_2 = (P_2 * T_1) / P_1 ) Substitute values: ( T_2 = (34 * 190) / 9 ) Calculate: ( T_2 \approx 721.11 , 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.