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If #7/4 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 #5/12 L#?

Answer 1

Elena Albarran

#P_2=67.2" "kPa#

#V_1=7/4L# #P_1=16kPa# #V_2=5/12L# #P_2=?#

#P_1*V_1=P_2*V_2" Boyle's law"#

#16*7/cancel(4)=P_2*5/cancel(12)#

#16*7=P_2*5/3#

#P_2=(16*7*3)/5#

#P_2=67.2" "kPa#

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Answer 2

Nicholas Agrusa

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

P1 * V1 = P2 * V2

(16 kPa) * (7/4 L) = P2 * (5/12 L)

P2 = (16 kPa * 7/4 L) / (5/12 L) P2 = (16 kPa * 7/4 L) * (12/5 L) P2 = 38.4 kPa

The gas will exert a pressure of 38.4 kPa when the container's volume changes to 5/12 L.

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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.