# If #14/3 L# of a gas at room temperature exerts a pressure of #45 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12/5 L#?

The pressure is

Boyle's Law is employed.

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Using Boyle's Law, we can solve for the new pressure:

P1 * V1 = P2 * V2

Where: P1 = initial pressure = 45 kPa V1 = initial volume = 14/3 L P2 = final pressure (what we're solving for) V2 = final volume = 12/5 L

Substituting the given values:

45 kPa * (14/3 L) = P2 * (12/5 L)

Solving for P2:

P2 = (45 kPa * (14/3 L)) / (12/5 L) P2 = (45 * 14) / (3 * 12/5) P2 = (630) / (36/5) P2 = 875 kPa

So, the gas will exert a pressure of 875 kPa when the container's volume changes to 12/5 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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