A gas occupies 12.3 at a pressure of 40.0 KPa. What is the volume when the pressure is increased to 60.0 KPa?

1 Answer

The volume is =8.2u, whatever the chosen units were to begin with...

Explanation:

We apply Boyle's Law at constant temperature and constant number of "mol"s:

P_1V_1=P_2V_2

The initial pressure is P_1=40 kPa

The initial volume is V_1=12.3

The final pressure is 60kPa

The final volume is

V_2=P_1/P_2*V_1

=40/60*12.3=8.2u