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A container with a volume of #14 L# contains a gas with a temperature of #120^o K#. If the temperature of the gas changes to #75 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Adrian Ayala

The volume is #=8.75L#

Charles' Law is employed.

#V_1/T_1=V_2/T_2#

#V_1=14L#

#T_1=120K#

#T_2=75K#

#V_2=V_1*T_2/T_1=14*75/120=8.75L#

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Answer 2

Nora Arzate

Using the combined gas law equation, (P_1V_1/T_1 = P_2V_2/T_2), where (P_1) and (P_2) are the initial and final pressures (which remain constant), (V_1) and (V_2) are the initial and final volumes, and (T_1) and (T_2) are the initial and final temperatures respectively, we can solve for (V_2):

[V_2 = \frac{P_1V_1T_2}{P_2T_1}]

Given: (V_1 = 14 , \text{L}), (T_1 = 120 , \text{K}), (T_2 = 75 , \text{K})

Plugging in the values:

[V_2 = \frac{14 , \text{L} \times 75 , \text{K}}{120 , \text{K}}]

[V_2 = \frac{1050 , \text{L}\cdot\text{K}}{120 , \text{K}}]

[V_2 = 8.75 , \text{L}]

Therefore, the container's new volume must be (8.75 , \text{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.