Question

A `780*mL` volume of gas `1.00*atm` pressure, and `310*K` temperature, is cooled to `295*K`, and `65*mm*Hg` pressure. What is the new volume?

Answer 1

For a given quantity of gas, .........`V_2=8.70*L.`

Explanation:

For a given quantity of gas, the combined gas law holds that `(P_1V_1)/T_1=(P_2V_2)/T_2`, temperature quoted in `"degrees Kelvin"`.

Thus `V_2=(P_1V_1)/T_1xxT_2/P_2`

`=(1.00*atmxx780*mL)/(310*K)xx(295*K)/((65*mm*Hg)/(760*mm*Hg*atm^-1)`

`=8.69xx10^3*mL=8.70*L`

We use here the useful relationship that `760*mm*Hg-=1*atm`.

And thus a measurement of length on a mercury manometer, which is fairly easy to do in a lab, can be used to solve Gas law problems.

Please note that I used `22` `""^@C`, `-=` `295*K`, and `37` `""^@C`, `-=` `310*K`. You can change the calculation accordingly if I have got your starting conditions wrong.