Problems and Solutions on Thermodynamics and Statistical Mechanics

Therrnodymm'ca 51

1055
A mixture of 0.1 mole of helium (71 = C,/Cv = 5/3) with 0.2 mole of
nitrogen (72 = 7/5), considered an ideal mixture of two ideal gases, is ini-
tially at 300K and occupies 4 litres. Show that the changes of temperature
and pressure of the system which occur when the gas is compressed slowly
and adiabatically can be described in terms of some intermediate value of

7. Calculate the magnitude of these changes when the volume is reduced
   by 1%.
   (UC, Berkeley)
   Solution:
   The entropy change for an ideal gas is

AS = nC, ln(Tf/z) + nR ln(Vf/x) ,

where n is the mole number, i and f indicate initial and final states respec-
tively. As the process is adiabatic the total entropy change in the nitrogen
gas and helium gas must be zero, that is, AS, + AS2 = 0. The expression
for AS then gives

where

Together with the equation of state for ideal gas, it gives

where

nlCpl+ n2Cp2

nlCvl+ n2G2 *

7=

Helium is monatomic, so that C,, = 3R/2, C,, = 5R/2; nitrogen is di-

atomic, so that C,, = 5R/2, C,, = 712. Consequently, 7 = 1.46.

When Vf = 0.99x, we have

Tf = 1.006Ti = 302 K ,

pf = 1.016pi = l.O16nRT/V = 2.0 x lo5 N/m2