If #9 L# of a gas at room temperature exerts a pressure of #15 kPa# on its container, what pressure will the gas exert if the container's volume changes to #4 L#?

1 Answer
Mar 6, 2016

Answer:

#33.75kPa#

Explanation:

From the information given for this question, we can see that this kind of situation is involving Boyle's Law.

Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .

In which from the definition, the equation is derived as;

#Pquadpropquad 1/V# or #P=k/V# or #PV=k#

#P=# Pressure of gas
#V=# Volume of gas
#k=# Constant

When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;

#P_iV_i=P_fV_f#

From the information given in this question;

#P_i=# Initial pressure of gas #=15kPa#
#P_f=# Final pressure of gas #=?kPa#
#V_i=# Initial volume of gas #=9L#
#V_f=# Final volume of gas #=4L#

Calculating #P_f#;

#(15kPa)(9L)=P_(f)(4L)#

#P_f=(135kPacancel(L))/(4cancel(L))#

#P_f=33.75kPa#