If #9/7 L# of a gas at room temperature exerts a pressure of #12 kPa# on its container, what pressure will the gas exert if the container's volume changes to #3/4 L#?

1 Answer
Mar 6, 2016

Answer:

Gas will exert a pressure of #20.47kPa#

Explanation:

Pressure #P#, volume #V# and temperature #T# (in Kelvin) have the relation #PV/T=k# where #k# is a constant.

As no change in the temperature has been mentioned it is assumed to be constant at room temperature and then #P_1xxV_1=P_2xxV_2#

In the given case, #9/7# #L# of gas inside of a container exerts a pressure of #12kPa# of pressure and is at a temperature of #30^oK#. When the volume of the gas changes to #3/4# #L#, the pressure becomes #P_2#, we should have

#12xx9/7=P_2xx3/4# or #P_2=(12xx9xx4)/(3xx7)=144/7=20.47#

i.e. #P_2=20.47kPa#