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A container with a volume of #5 L# contains a gas with a temperature of #320^o K#. If the temperature of the gas changes to #280^o K# without any change in pressure, what must the container's new volume be?

Answer 1

William Audy

The new volume is #=4.375L#

Charles' Law is used.

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Answer 2

Leo Abeyta

Using the ideal gas law formula, (PV = nRT), where (P) is pressure (which remains constant), (V) is volume, (n) is the number of moles of gas (which remains constant), (R) is the ideal gas constant, and (T) is temperature in Kelvin:

(V_1 / T_1 = V_2 / T_2)

Given: (V_1 = 5) L (T_1 = 320) K (T_2 = 280) K

(V_2 = (V_1 \cdot T_2) / T_1)

(V_2 = (5 \cdot 280) / 320)

(V_2 = 4.375) L

The container's new volume must be approximately 4.375 liters.

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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.