A #2.94L# volume of gas at #294K# and under #69.6kPa# pressure enclosed in a piston, is cooled to #257K# and the pressure reduced to #35.6*kPa#. What is the new volume?

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Answer 1 · 2017-01-23T18:20:38

#V_2~=5*L#

#(P_1V_1)/T_1=(P_2V_2)/T_2#, from the #"combined gas law"#.

And thus #V_2=(P_1V_1)/T_1xxT_2/P_2#, which expression CLEARLY has the units of volume.

So #V_2=(69.9*cancel(kPa)xx2.94*L)/(294*cancel(K))xx(257*cancel(K))/(35.6*cancel(kPa))=??L#

A #2.94*L# volume of gas at #294*K# and under #69.6*kPa# pressure enclosed in a piston, is cooled to #257*K# and the pressure reduced to #35.6*kPa#. What is the new volume?