If #7/5# #L# of a gas at room temperature exerts a pressure of #3# #kPa# on its container, what pressure will the gas exert if the container's volume changes to #5/3# #L#?

1 Answer
Feb 19, 2016

Answer:

Start with the Combined Gas Law: #(P_1V_1)/T_1=(P_2V_2)/T_2#

Temperature is constant, so #(P_1V_1)=(P_2V_2) to P_2=(P_1V_1)/V_2 = (3*(7/5))/(5/3) = (21/5)/(5/3)=36/25# #kPa#

Explanation:

We could memorise Boyle's, Charles' and Gay-Lussac's Laws, which relate various combinations of pressure, volume and temperature, but I find it simpler to just memorise the Combined Gas Law and then eliminate the thing that is kept constant.

#(P_1V_1)/T_1=(P_2V_2)/T_2#

In this case, the temperature is kept constant, so we can ignore it:

#(P_1V_1)=(P_2V_2)#

Rearranging:

#P_2=(P_1V_1)/V_2#