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A container with a volume of #48 L# contains a gas with a temperature of #210^o K#. If the temperature of the gas changes to #690 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Adrian Ayala

The volume is #=157.7#

We use Charles' Law.

#V_1/T_1=V_2/T_2#

#V_1=48 L#

#T_1=210K#

#T_2=690K#

#V_2=(V_1T_2)/T_2#

#=48*690/210=157.7L#

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Answer 2

Nora Arzate

Use the combined gas law formula: ( \frac{V_1}{T_1} = \frac{V_2}{T_2} ). Rearrange the formula to solve for ( V_2 ): ( V_2 = \frac{V_1 \times T_2}{T_1} ). Substitute the given values: ( V_2 = \frac{48 \times 690}{210} = 157.14 , \text{L} ). Therefore, the new volume of the container must be 157.14 L.

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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.