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?

1 Answer
Nov 14, 2016

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.