If #18 L# of a gas at room temperature exerts a pressure of #64 kPa# on its container, what pressure will the gas exert if the container's volume changes to #24 L#?

Answer 1

#P_2 = 64 *18/24 = 3/4*64 = 48 kPa #

Given: Container with initial, #V_1 = 18L and P_1=64kPa# Final #V_2= 24L# Required: New Pressure Solution Strategy: Straight application of Boyle's Law #P_1V_1=P_2V_2; P_2=P_1V_1/V_2# substituting the known values #P_2 = 64 *18/24 = 3/4*64 = 48 kPa #
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Answer 2

Using Boyle's Law, the initial and final pressure and volume can be related:

[ P_1 \cdot V_1 = P_2 \cdot V_2 ]

[ 64 , \text{kPa} \cdot 18 , \text{L} = P_2 \cdot 24 , \text{L} ]

[ P_2 = \frac{64 , \text{kPa} \cdot 18 , \text{L}}{24 , \text{L}} ]

[ P_2 = 48 , \text{kPa} ]

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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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