Question

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"`?

Answer 1

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.