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

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.