A container with a volume of #6 L# contains a gas with a temperature of #280^o K#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?
The new volume will be
We start with the ideal gas law:
and move all of the constant terms to one side. We know from the question that the temperature and the volume are changing - everything else is constant, in other words:
This means that we can set the two different sets of volumes and temperatures equal to each other using this equation, since we know they are constant:
We can now solve for the new volume we are trying to find and plug in the numbers from the question:
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Using the combined gas law ( PV/T = k ), where ( P ) is pressure, ( V ) is volume, and ( T ) is temperature, and since pressure remains constant, the equation simplifies to ( V_1/T_1 = V_2/T_2 ). Substituting the given values, ( V_1 = 6 ) L and ( T_1 = 280 ) K, and solving for ( V_2 ) with ( T_2 = 320 ) K:
( V_1/T_1 = V_2/T_2 ) ( 6/280 = V_2/320 ) ( V_2 = (6 \times 320)/280 ) ( V_2 = 6.857 ) L
Therefore, the container's new volume must be approximately 6.857 liters.
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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.
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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