# If #9/4 L# of a gas at room temperature exerts a pressure of #16 kPa# on its container, what pressure will the gas exert if the container's volume changes to #9/7 L#?

The pressure is

Using Boyle's Law

The last bit of pressure is

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To find the new pressure, use Boyle's law:

[P_1V_1 = P_2V_2]

Given: (P_1 = 16 , kPa) (V_1 = \frac{9}{4} , L) (V_2 = \frac{9}{7} , L)

Substitute the given values into the equation and solve for (P_2):

[P_2 = \frac{P_1V_1}{V_2}]

[P_2 = \frac{(16 , kPa) \times \left(\frac{9}{4} , L\right)}{\frac{9}{7} , L}]

[P_2 = \frac{144 , kPa}{\frac{9}{7}}]

[P_2 = \frac{144 , kPa \times 7}{9}]

[P_2 = \frac{1008 , kPa}{9}]

[P_2 = 112 , kPa]

So, the gas will exert a pressure of (112 , kPa) if the container's volume changes to (\frac{9}{7} , 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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