A constant temperature, a #12*L# volume of gas in a piston, is compressed to #3*L#. If the original pressure was #41*kPa#, what is the final pressure?

2 Answers
Aug 11, 2017

Answer:

The pressure is #=164kPa#

Explanation:

We apply Boyle's Law

#P_1V_1=P_2V_2#, the temperature is constant

The initial pressure is #P_1=41kPa#

The initial volume is #V_1=12L#

The final volume is #V_2=3L#

The final pressure is

#P_2=V_1/V_2*P_1=12/3*41=164kPa#

Aug 11, 2017

Answer:

Well, we use old Boyle's law.........and get #P_2=164*kPa#.

Explanation:

At constant temperature, and constant amount of gas, the product #PxxV=k#, where #k# is some constant....

And thus, if we solve for #k# at different conditions of volume and pressure, then #P_1V_1=P_2V_2#.

The utility of this equation is that we can use whatever whack units of volume and pressure we like, #"pints, pounds per square inch, torr, atmospheres, quarts"#, as long as we are consistent.

And so...#P_2=(P_1V_1)/V_2=(41*kPaxx12*L)/(3*L)=164*kPa#