If the volume of an ideal gas at #27^@ "C"# halves, what is its new temperature? Why can I not use the celsius temperature to do this?
1 Answer
Assuming an ideal gas and that we are at constant pressure and mols of ideal gas, we construct two ideal gas states:
#PV_1 = nRT_1#
#PV_2 = nRT_2#
Solving for
#V_1/T_1 = V_2/T_2# which is Charles law, relating volume and temperature changes at constant pressure and mols of gas .
Designate the original volume as
#T_2 = (V_2/V_1) T_1#
#= (cancelV)/(0.5cancel(V)) (27^@ "C")#
#=> T_2 = ul(color(red)(54^@ "C"))#
which is completely incorrect... What we should have done is recognize that
#color(blue)(T_2) = 1/0.5("300.15 K")#
#=# #color(blue)(6.0_(03) xx 10^2 "K")#
or
(Obviously, the celsius scale has destroyed the meaning of the logic; we do not see a clear-cut doubling of the celsius temperature.)