A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?

1 Answer
May 5, 2016

Answer:

#V_2=31.7L#

Explanation:

Let us first list the given data:

Position 1:
STP means that #P=1atm=760mmHg# and #T=273K#.
The volume occupied by the gas is #V=12.0L#.

Position 2:
The new conditions are:
#P=575mmHg# and #T=2xx273=546K#.
#V=?#

To find the volume of the gas at position 2, we can rearrange the ideal gas law: #PV=nRT# to be:

#(PV)/T=k# where #k=nR#, which is constant.

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

#=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(760cancel(mmHg)xx12.0L)/(273cancel(K))xx(546cancel(K))/(575cancel(mmHg))=31.7L#