# A container with a volume of #32 L# contains a gas with a temperature of #480^o K#. If the temperature of the gas changes to #160 ^o K# without any change in pressure, what must the container's new volume be?

The new volume is

We apply Charles' Law

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Using the combined gas law, ( \frac{{V_1}}{{T_1}} = \frac{{V_2}}{{T_2}} ), where ( V_1 = 32 , \text{L} ), ( T_1 = 480 , \text{K} ), and ( T_2 = 160 , \text{K} ), we can solve for ( V_2 ). Thus, ( V_2 = \frac{{V_1 \cdot T_2}}{{T_1}} = \frac{{32 , \text{L} \cdot 160 , \text{K}}}{{480 , \text{K}}} = 10.67 , \text{L} ). Therefore, the container's new volume should be approximately ( 10.67 , \text{L} ).

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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.

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- The gas inside of a container exerts #15 Pa# of pressure and is at a temperature of #180 ^o K#. If the pressure in the container changes to #36 Pa# with no change in the container's volume, what is the new temperature of the gas?
- If #15 L# of a gas at room temperature exerts a pressure of #5 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12 L#?
- How do you measure fluid flow?

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