# A container with a volume of #4 L# contains a gas with a temperature of #770^o K#. If the temperature of the gas changes to #420^o K# without any change in pressure, what must the container's new volume be?

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Using the combined gas law formula ( \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} ), where ( P ) is pressure, ( V ) is volume, and ( T ) is temperature, we can solve for the new volume:

( \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} )

Given: ( V_1 = 4 , \text{L} ) ( T_1 = 770 , \text{K} ) ( T_2 = 420 , \text{K} ) ( P_1 = P_2 ) (pressure remains constant)

( \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} )

( \frac{P_1 \times 4}{770} = \frac{P_2 \times V_2}{420} )

( \frac{4P_1}{770} = \frac{P_2 \times V_2}{420} )

( \frac{4P_1 \times 420}{770} = P_2 \times V_2 )

( V_2 = \frac{4P_1 \times 420}{770} )

Since ( P_1 = P_2 ), we can replace ( P_1 ) with ( P_2 ):

( V_2 = \frac{4P_2 \times 420}{770} )

( V_2 = \frac{1680}{770} )

( V_2 ≈ 2.18 , \text{L} )

Therefore, the container's new volume should be approximately ( 2.18 , \text{L} ).

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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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