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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 #13/12 L#?

Answer 1

Olivia Askew

The pressure is #=25.85kPa#

Using Boyle's Law

#P_1V_1=P_2V_2#

#P_1=7/4L#

#V_1=16kPa#

#V_2=13/12L#

#P_2=(P_1V_1)/V_2#

#=(7/4*16)/(13/12)=28/13*12=25.85kPa#

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Answer 2

Sebastian Acosta

To find the new pressure, use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature is constant.

( P_1V_1 = P_2V_2 )

( P_2 = \frac{{P_1V_1}}{{V_2}} )

Substitute the given values:

( P_2 = \frac{{16 \text{ kPa} \times \frac{7}{4} \text{ L}}}{{\frac{13}{12} \text{ L}}} )

( P_2 = \frac{{16 \times 7}}{{4 \times \frac{13}{12}}} )

( P_2 = \frac{{16 \times 7 \times 12}}{{4 \times 13}} )

( P_2 = \frac{{1344}}{{52}} )

( P_2 ≈ 25.85 \text{ kPa} )

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