Question

one mole of a real gas is subjected to a process from (2 bar, 30 L, 300 K) to (2 bar, 40 L, 500 K)
Given : $C_v=25$ J/ mol K, $C_p=40$ J/ mol K
Calculate $△U$:

Answer 1

Given,
Initial pressure, $P_1=2bar$
Initial Volume, $V_1=30L$
Initial temperature $T_1=300K$
Again,
Final pressure, $P_2=2bar$
Final Volume, $V_2=40L$
Final temperature $T_2=500K$

According to first law of thermodynamics,
$\Delta U=q+W$

We know,
$q=C_p△T\times n=40\times [500-300]\times 1=40\times 200=8000J\therefore W=-P_{ext}\times △V\times 100$
(workdone)
$=-2\times (40-30)\times 100=-2\times 10\times 100J=-2000J\therefore △U=8000+(-2000)J△U=6000J$