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?

Answer 1

#V_2~=8,295 l#

#"change 460^o to Kelvin :" 460+273=733 K# #V_1/T_1=V_2/T_2# #19/733=V_2/320# #V_2=(320*19)/733# #V_2~=8,295 l#
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Answer 2

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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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.

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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