What is the new volume for a #50.4*mL# volume of gas, at #742*mm*Hg# pressure and #293.15*K#, for which conditions are changed to #"STP"#?

1 Answer
Sep 10, 2017

Answer:

Well, STP specifies a pressure of #10^5*Pa#, and a temperature of #273.15*K#.....

Explanation:

We use the old combined gas law, which says for a GIVEN amount of gas.....

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

And so we solve for #V_2=(P_1V_1T_2)/(P_2T_1)#, which product clearly has units of volume. Why?

And we plug in the numbers knowing that #100*kPa-=750*mm*Hg#...

#V_2=(742*mm*Hgxx50.4xx10^-3Lxx273.15*K)/(750*mm*Hgxx293.15*K)#

#V_2=0.0465*L-=46.5*cm^3#; i.e. the volume is SLIGHTLY compressed......

Note that while this does seem complicated, all I have done is use the appropriate units. That I got an answer in #cm^3# suggests that my order of operations was right.