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If #5/7 L# of a gas at room temperature exerts a pressure of #9 kPa# on its container, what pressure will the gas exert if the container's volume changes to #8/5 L#?

Answer 1

Lily Adams

#P_2~=4,02 kPa#

#P_1*V_1=P_2*V_2" Boyle-Mariotte's law"# #9*5/7=P_2*8/5# #P_2=(9*5*5)/(7*8)# #P_2~=4,02 kPa#

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Answer 2

Amelia Adams

To find the new pressure, use Boyle's law equation: ( P_1V_1 = P_2V_2 ). Plug in the given values: ( P_1 = 9 ) kPa, ( V_1 = \frac{5}{7} ) L, ( V_2 = \frac{8}{5} ) L. Solve for ( P_2 ).

( P_2 = \frac{P_1 \cdot V_1}{V_2} )

( P_2 = \frac{9 \times \frac{5}{7}}{\frac{8}{5}} )

( P_2 = \frac{45}{56} ) 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.