A container with a volume of #19 L# contains a gas with a temperature of #460^o C#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?
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To solve this problem, we'll use the combined gas law, which states:
[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} ]
Since the pressure (( P )) remains constant, we can simplify the equation to:
[ \frac{V_1}{T_1} = \frac{V_2}{T_2} ]
Given:
- ( V_1 = 19 , \text{L} ) (initial volume)
- ( T_1 = 460 , ^\circ \text{C} = 733 , \text{K} ) (initial temperature)
- ( T_2 = 320 , ^\circ \text{K} ) (final temperature)
We can rearrange the equation to solve for ( V_2 ):
[ V_2 = \frac{T_2 \times V_1}{T_1} ]
Substituting the values:
[ V_2 = \frac{320 , \text{K} \times 19 , \text{L}}{733 , \text{K}} ]
[ V_2 \approx 8.29 , \text{L} ]
So, the container's new volume should be approximately 8.29 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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