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
anor277
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.

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