If #3 L# of a gas at room temperature exerts a pressure of #5 kPa# on its container, what pressure will the gas exert if it the container's volume changes to #2/3 L#?

Answer 1

William Audy

The pressure is #=22.5kPa#

Using Boyle's Law

#P_1V_1=P_2V_2#

#P_1=5kPa#

#V_1=3L#

#V_2=2/3L#

#P_2=V_1/V_2*P_1#

#=3/(2/3)*5=45/2=22.5kPa#

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Answer 2

Charles Aeschliman

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

[ P_1 \times V_1 = P_2 \times V_2 ]

Given:

Initial pressure ((P_1)) = 5 kPa
Initial volume ((V_1)) = 3 L
Final volume ((V_2)) = ( \frac{2}{3} ) L

Substitute the values into the equation:

[ 5 \times 3 = P_2 \times \frac{2}{3} ]

[ P_2 = \frac{5 \times 3}{\frac{2}{3}} ]

[ P_2 = \frac{15}{\frac{2}{3}} ]

[ P_2 = 22.5 , \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.