The gas inside of a container exerts #48 Pa# of pressure and is at a temperature of #320 ^o K#. If the pressure in the container changes to #64 Pa# with no change in the container's volume, what is the new temperature of the gas?

Answer 1

Layla Ahmed

The new temperature is #=426.7K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=48Pa#

The initial temperature is #T_1=320K#

The final pressure is #P_2=64Pa#

The final temperature is

#T_2=P_2/P_1*T_1#

#=64/48*320#

#=426.7K#

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Answer 2

Adrian Ayala

use gay-lussac's law .

gay-lusaac's law gives you the relation between pressure and temperature of a gas inside a container with a fixed volume : # P_1/T_1 = P_2/T_2 # #P_1 # is the pressure exerted by a gas at temperature #T_1# and #P_2# is the pressure exerted by the gas at temperature # T_2 # . now, substituting the values : #48/320 =64/T # # T= 426.67 k #

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Answer 3

Parker Abbott

The new temperature of the gas is 426.67 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.