The gas inside of a container exerts #12 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

#T_2=1706,67 ^o K#

#P_1/T_1=P_2/T_2#
#P_1=12Pa# #T_1=320 ^o K#
#P_2=64 Pa# #T_2=?#
#12/320=64/T_2#
#T_2=(320*64)/12#
#T_2=1706,67 ^o K#
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Answer 2

To find the new temperature of the gas, you can use the formula: ( P_1 \cdot T_1 = P_2 \cdot T_2 ), where ( P_1 ) and ( T_1 ) are the initial pressure and temperature, and ( P_2 ) and ( T_2 ) are the final pressure and temperature.

Given: ( P_1 = 12 , Pa ) ( T_1 = 320 , K ) ( P_2 = 64 , Pa )

Plug in the values: ( 12 , Pa \cdot 320 , K = 64 , Pa \cdot T_2 )

Solve for ( T_2 ): ( T_2 = \frac{12 , Pa \cdot 320 , K}{64 , Pa} )

( T_2 = 60 , 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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