Question

Consider a reversible isobaric process from state I (P, Vi , Ti) to state II (P, Vf , Tf ), which we call “path I”. The “path I” is a single step process and, therefore, pressure is held constant during the entire process. “Path II”, on the other hand, is also isobaric overall, but involves two steps: reversible isochore (step 1) + reversible isothermal (step 2): (Ti , Vi) → (Tf , Vi) → (Tf , Vf ). Assume that the gas is 1.00 mol ideal gas. (a) Express the amount of work (ω) done during each path in terms of temperature ratio, x = Ti Tf . (b) Which path generated more work? You may plot w as a function of x and see which one is larger. (c) Calculate ∆U and q for two thermodynamic paths.

Answer 1

Isobaric process are those processes in which pressure remains the same.

In the path 1 we have Pressure = P = constant, initial temperature = Ti, final temperature = Tf, initial volume = Vi, final volume = Vf

workdone is given by w = -Pext $\Delta$ V where $\Delta$ V = Vf-Vi

So, w = -P $\Delta$ V = -P(Vf-Vi)

By ideal gas equation PV = nRT where n is number of moles = 1 (given)

we have PV = RT

so, for V = Vi, T =Ti we get Vi = RTi/P similarly Vf = RTf/P

so, putting these values in work equation we get w = -P(RTf/P - RTi/P) = -R(Tf-Ti)

Taking Tf common we get w = -RTf(1-Ti/Tf) = -RTf(1-x) as x = Ti/Tf

Change in internal energy of this process = $\Delta$ U = nCv $\Delta$ T where Cv is the molar heat capacity of ideal gas = 3R/2 , n is the number of moles = 1 , $\Delta$ T = change in temperature = Tf-Ti

so, $\Delta$ U = 3R*(Tf-Ti)/2

from the first law of thermodyanmics which states that the total energy of an isolated system is constant we have $\Delta$ U = q+w so, by putting the values of $\Delta$ U and w we get q = $\Delta$ U-w

so, q = 3R*(Tf-Ti)/2 - (R(Tf-Ti)) = R*(Tf-Ti)/2

For path 2 we have two steps. First one is isochoric (for isochoric processes the volume is constant) and isothermal (for isothermal processes the temperature is constant)

so, for step 1. we have workdone w1 = -Pext $\Delta$ V = 0 as $\Delta$ V = 0 (volume is constant)

for step 2, we have workdone w2 = -Pext $\Delta$ V = -P(Vf-Vi) where initial volume = Vi, final volume = Vf and P is the final pressure

P = RTf/Vf as n=1 from ideal gas equation. so, w2 = -RTf/Vf

as step 2 is isothermal we also have Pint*Vi = PfVf -- Eq1(For isothermal PV = constant , Pint is the intermediate pressure)

as step 1 is isochoric we also have Pi/Ti = Pint/Tf ( Final pressure is Pf and as the overall process is isobaric Pi = Pf so, we have Pf/Ti = Pint/Tf) --- Eq2

From Eq1 and Eq2 we get PfTf/Ti = PfVf/Vi so, Tf/Ti = Vf/Vi so, Vi/Vf = Ti/Tf

wo, wtotal = w1 + w2 = 0 + (-RTf/Vf(Vf-Vi)) = -RTf(1-Vi/Vf) so, wtotal = -RTf(1-Ti/Tf) = -RTf(1-x) as x = Ti/Tf

As the workdone in both the paths are same we can say that work done in both the paths are the same.

As change in internal energy is only dependent on the initial and final temperature. We have -

$\Delta$ U = nCv $\Delta$ T where Cv is the molar heat capacity of ideal gas = 3R/2 , n is the number of moles = 1 $\Delta$ T = change in temperature = Tf-Ti

so, $\Delta$ U = 3R*(Tf-Ti)/2

from the first law of thermodyanmics which states that the total energy of an isolated system is constant we have $\Delta$ U = q+w so, by putting the values of $\Delta$ U and w we get q = $\Delta$ U-w

so, q = 3R*(Tf-Ti)/2 - (R(Tf-Ti)) = R*(Tf-Ti)/2