If #7/6 L# of a gas at room temperature exerts a pressure of #2 kPa# on its container, what pressure will the gas exert if the container's volume changes to #5/4 L#?

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Answer 1 · 2018-02-24T06:57:29

The pressure will be #=1.87kPa#

Apply Boyle's Law

#P_1V_1=P_2V_2#, at constant temperature

The initial pressure is #P_1=2kPa#

The initial volume is #V_1=7/6L#

The final volume is #V_2=5/4L#

The final pressure is

#P_2=V_1/V_2*P_1=(7/6)/(5/4)*2=1.87kPa#