If #18 L# of a gas at room temperature exerts a pressure of #5 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12 L#?

Answer 1

This a pretty straight forward application of Boyle's Law, expressed by the equation: #P_1V_1 = P_2V_2#
#18*5 = 12*V_2 " solve for " V_2 #
#v_2 =5 18/12 = 5 3/2 =15/2 =7.5 kPa#

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Answer 2

Nicholas Agrusa

Using the combined gas law equation, we have:

(P1 * V1) / T1 = (P2 * V2) / T2

Where: P1 = initial pressure V1 = initial volume T1 = initial temperature (assumed to remain constant as "room temperature") P2 = final pressure V2 = final volume T2 = final temperature (assumed to remain constant as "room temperature")

Given: P1 = 5 kPa V1 = 18 L V2 = 12 L T1 = T2 (room temperature assumed constant)

Solving for P2: (P1 * V1) / V2 = P2

Substituting the values: P2 = (5 kPa * 18 L) / 12 L

P2 = 7.5 kPa

Therefore, if the container's volume changes to 12 L, the gas will exert a pressure of 7.5 kPa at room temperature.

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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.