If #10/3 L# of a gas at room temperature exerts a pressure of #42 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12/7 L#?

2 Answers
Mar 21, 2018

Answer:

The new pressure is #=81.7kPa#

Explanation:

Apply Boyle's Law

#"Pressure" xx"Volume"="Constant"# at constant temperature

#P_1V_1=P_2V_2#

The initial pressure is #P_1=42kPa#

The initial volume is #V_1=10/3L#

The final volume is #V_2=12/7L#

The final pressure is

#P_2=V_1/V_2*P_1=(10/3)/(12/7)*42=81.7kPa#

Mar 22, 2018

Answer:

#81.7 \ "kPa"#

Explanation:

We use Boyle's law, if the volume and the number of moles is constant, and we get,

#Pprop1/V# or #P_1V_1=P_2V_2#

Now, we plug in the given values, and we get

#42 \ "kPa"*10/3color(red)cancelcolor(black)"L"=P_2*12/7color(red)cancelcolor(black)"L"#

#P_2~~81.7 \ "kPa"#

So, the new pressure is approximately #81.7# kilopascals.