At constant temperature, a 3.8*L3.8⋅L volume of gas at 765*mm*Hg765⋅mm⋅Hg, was expanded to give a pressure of 0.500*atm0.500⋅atm. What is the new volume?

1 Answer
anor277
May 4, 2017

P_1V_1=P_2V_2P1V1=P2V2 under conditions of constant temperature.

Explanation:

We need to know that 1*atm1⋅atm will support a column of mercury that is 760*mm760⋅mm high, and thus a mercury column may used as a highly visual measurement of pressure:

1*atm-=760*mm*Hg1⋅atm≡760⋅mm⋅Hg (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......

P_1=(765*mm*Hg)/(760*mm*Hg*atm^-1)=1.01*atm.P1=765⋅mm⋅Hg760⋅mm⋅Hg⋅atm−1=1.01⋅atm.

We solve for V_2=(P_1V_1)/P_2=(1.01*atmxx3.8*L)/(0.500*atm)=7.60*LV2=P1V1P2=1.01⋅atm×3.8⋅L0.500⋅atm=7.60⋅L

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