If for an ideal gas its volume is compressed by a factor of 3, and the gas has cooled down as a result, by what factor is the pressure changed?

Question

Emily Abbate · Accepted Answer

I'm getting that it's more than three times.

First of all, here are our assumptions:

We are told that #V_2 = 1/3V_1#, and we infer that #DeltaT ne 0# and #DeltaP ne 0#. The reversible work is

#w_"rev" = -int_(V_1)^(V_2)PdV#,

which is the cause for #DeltaT ne 0#. For this compression, work is done on the gas, which means #w > 0# and #DeltaV < 0#.

Also, by compressing the gas, we conceptually have that the particles move closer together and the gas cools down; since

it must decrease in temperature. Mathematically:

#Delta(PV) = nRDeltaT#

Since #Vdarr#, if we had #DeltaP = 0#, then

#PDeltaV + cancel(VDeltaP)^(0) = nRDeltaT#,

and #Tdarr#.

If we did have #DeltaT = 0# but #DeltaP ne 0#, then

#P_1V_1 = P_2V_2#, and #P_1/P_2 prop V_2/V_1#,

i.e. dividing the volume by three means tripling the pressure.

In comparison, since the heat flow cannot escape the adiabatic container, the additional temperature change must have gone into compression in an adiabatic enclosure. So, the more the temperature decreases , the more the volume decreases, and the more the pressure must increase to balance out the additional volume decrease.

Or, mathematically:

#V_1/V_2 prop T_1/T_2# and #P_2/P_1 prop V_1/V_2#

meaning that as #Tdarr#, #Vdarr# and #Puarr# relative to the case where #DeltaT = 0# (where the new pressure would have been three times as large).

Therefore, since the pressure would have tripled if #DeltaT = 0#, but #DeltaT < 0#, the change in pressure is more than three times.

Benjamin Asiedu · Answer

undefined The pressure is increased by a factor of 9.