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The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #450 ^o K#. If the pressure in the container changes to #27 Pa# with no change in the container's volume, what is the new temperature of the gas?

Answer 1

Charlotte Aaron

The temperature is #=675K#

We apply the Gay Lussac Law.

#P_1/T_1=P_2/T_2#

#P_1=18Pa#

#T_1=450K#

#P_2=27Pa#

#T_2=P_2/P_1*T_1#

#=27/18*450=675K#

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Answer 2

Xavier Adame

Using the ideal gas law ( PV = nRT ), where ( P ) is pressure, ( V ) is volume, ( n ) is the number of moles, ( R ) is the gas constant, and ( T ) is temperature, we can rearrange the equation to solve for temperature: ( T = \frac{{PV}}{{nR}} ). Since the volume and the number of moles remain constant, the new temperature (( T_{new} )) can be calculated using the initial pressure (( P_{initial} )) and the final pressure (( P_{final} )): [ T_{new} = \frac{{P_{final} \times T_{initial}}}{{P_{initial}}} ] [ T_{new} = \frac{{27 , Pa \times 450 , K}}{{18 , Pa}} ] [ T_{new} = \frac{{12150 , K \cdot Pa}}{{18 , Pa}} ] [ T_{new} = 675 , K ]

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Answer from HIX Tutor

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