Question

one mole of an ideal monoatomic gas is taken through the cycle shown in the figure. the process A:B is a reversible isothermal expansion. calculate the net work done by the gas?

Answer 1

Solution;

Work done in process AB;

$W_{AB}=-P_AV_Aln(\frac{V_B}{V_A})$

$W_{AB}=-5×1.031×10^5×10×10^{-3}×ln(\frac{50}{10})$

$W_{AB}=-8151.80J$

Work done in process BC;

$W_{BC}=-P_B(V_A-V_B)$

$W_{BC}=-1×1.031×10^5(10-50)×10^{-3}$

$W_{BC}=4124J$

Work done in process CA;

$W_{CA}=0$

Therefore,the net work done is;

$W_{net}=Q_{AB}+Q_{BC}+Q_{CA}$

$W_{net}=8151.80+4124+0$

$W_{net}=12275.8J$