Question

Exercise: A system consists of He gas with an internal' energy of 1500 J. If the gas receives 525 J of heat from its surroundings and does PV work on them (P = 1.0 atm) at same time, what is its final internal energy? Wsys = -PAV = -1.00 atm*(8.23 -4.00)L = -4.23 L-atm Einit = 1500 J -4.23 L-atm x 101.3J/Latm = -428J 48.23 L AE = Efinal - Einitial = qsys + Wsys = +525J – 428J AE = +97J He He , 4.00 Lt qsys=+525J +97J = Efinal - 1500J Efinal = 1500J + 97J = 1597J

If someone could please explain where the (8.23-4.00) comes from I really dont understand that . Thanks!

Answer 1

It is example of expansion (PV) work, it is under constant external pressure (1 atm) , thus it is non-reversible work. We cannot use integrated equation that is applicable for reversible expansion.

Work done by system in Non-reversible expansion is given by ,

W = -P_ex. * $\DeltaV$$\Delta V$

$\DeltaV$$\Delta V$ , can be calculated by volumes at initial and final states.

W = - P_ext.* (V_f - V_i)

V_f = Final volume after expansion

V_i = Initial volume before expansion

For this problem volume changes are represented in diagram and their values are indicated in diagram ( nothing else mentioned regarding volume, for this problem volumes are not calculated but already provided with : no need of any calculation).

So, $\Delta$ V = (V_f - V_i) = (8.23 - 4.00 ) L