Question

Two moles of helium gas undergo a cyclic process as shown
in figure. Assuming the gas to be ideal, calculate the
following quantities in this process.
(a) The net change in the heat energy.
(b) The net work done.
(c) The net change in internal energy.

Answer 1

Given number of moles n = 2
Process AB and CD are isobaric.
Hence, Q$_{AB} = - Q_{CD}$
[Because ($\Delta T)_{AB}$ = + 100K whereas ($\Delta T )_{CD}$ =-100 K and
$Q_{isobaric} =nC_p \Delta T]$
or $ \, \, \, \, \, \, \, \, \, \, \, \, \, Q_{AB}+Q_{CD}=0$
Process BC is isothermal ($\Delta$U = 0)
$\therefore \, \, \, \, \, \, \, \, \, Q_{BC}=W_{BC}=nRT_B ln \bigg(\frac{P_B}{P_C}\bigg)$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, =(2)(8.34)(400)ln \bigg(\frac{2}{1}\bigg)$
$ \, \, \, \, \, \, \, \, \, Q_{BC}=4608 J$
Similarly, process DA is also isothermal hence,
$Q_{DA}=Q_{DA}=nRT_D ln \bigg(\frac{p_D}{p_A}\bigg) =(2)(8.31)(300)ln \bigg(\frac{1}{2}\bigg)$
$Q_{BC}=-3456 J$
(a) Net heat exchange in the process
$Q=Q_{AB}+Q_{BC}+Q_{CD}+Q_{DA}=(4608-3456) J$
Q=1152 J
(b) From first law of thermodynamics, AU = 0
(in complete cycle) Q$_{net}=W_{net}$
Hence, W =Q= 1152 J
(c) Since, $T_i =T_f $ therefore net change in internal energy,
dU = 0.