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If #8 L# of a gas at room temperature exerts a pressure of #32 kPa# on its container, what pressure will the gas exert if the container's volume changes to #4 L#?

Answer 1

Emily Abbate

#P_2=64" " kPa#

#"use the Boyle's law"# #P_1*V_1=P_2*V_2# #32*cancel(8)=P_2*cancel(4)# #P_2=2*32# #P_2=64" " kPa#

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Answer 2

Sebastian Acosta

According to Boyle's Law, the pressure and volume of a gas are inversely proportional when the temperature is constant. Using the formula P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume:

P1V1 = P2V2

(32 kPa)(8 L) = P2(4 L)

Solving for P2:

P2 = (32 kPa * 8 L) / 4 L P2 = 64 kPa

Therefore, if the container's volume changes to 4 L, the gas will exert a pressure of 64 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.