Question

A system containing 2 moles of CO 2 (g) is initially at 25 0 C and 10atm and is confined to a cylinder that has a cross section of 10 cm 2 . The system expands adiabatically against an external pressure of 1 atm until the piston has moved upwards through 20cm. Assume that CO 2 is an ideal gas with C v,m = 28.8 J/K mol. Calculate q, w, Δ U, Δ T and Δ S.

Answer 1

Control mass: carbon dioxide confined inside the cylinder.
Process: constant pressure adiabatic process since both temperature and volume change.
Model: assuming carbon dioxide behaves like an ideal gas.
Analysis: since the process is adiabatic there is no heat transfer has taken place. also the change in kinetic and potential energy are very negligible.

In adiabatic process w = Δ U as q is 0 because of constant heat.

The work done in this problem is the boundary work against the atmospheric pressure Po = 100kPa(= 1atm).
the change in volume be = 10cm2 * 20cm = 200cm3 = 0.0002m3.
therefore work done by the system w = Po * ΔV = 100kPa * 0.0002m3 = 0.02kJ = 20Joule = w
ΔU = 20 Joule.
q = 0.
using the relation PV = mRT where P = 10atm = 1000kPa; R = 0.189J/gm.K; T = 25 degree C = 298K; m = 88gm. we find 0.00496m3.
final volume is = 0.00496 + 0.0002m3 = 0.00516m3.
using the relation V1 / T1 = V2 / T2 we find the final temperature is = 310K.
ΔT = (310 -298)K = 12K
we know that Cp,m - Cv,m = R or, Cp,m = R + Cv,m = 0.842 J/gm-K.
for an ideal gas the change in entropy is given by s2 - s1 = Cp,m * ln(T2/T1) - R * ln(P2/P1)
putting the values we get 0.03324J/gm-K. Therefore for 88gm of CO2 the entropy change is Δ S = 2.92512 JK^-1.