Question

Figure shows a cyclic process $ABCDBEA$ performed on an ideal cycle. If $P_A=2atm$, $P_B=5atm$ and $P_6=6atm$, $V_E-V_A=20\text{litre}$, Find the work done by the gas in the complete process $(1atm=1\times {10}^{5}Pa)$

• A. $3.4kJ$
• B. $2.67kJ$
• C. $1.33kJ$
• D. $4.25kJ$

Answer 1

The correct option is B $2.67kJ$

The complete cyclic process can be visualized as made up of two cycles,
i.e., cycle $AEBA$ (clockwise) and cycle $BDCB$ (counter-clockwise).
Work done by the gas during the cycle $AEBA$ should be positive,
$W_1$= Area of the loop in the $P-V$ diagram.
$W_1=\frac{1}{2}\left(\text{base}\right)\left(\text{altitude}\right)$
$=\frac{1}{2}(V_E-V_A)(P_B-P_A)$

$=\frac{1}{2}(20\times {10}^{-3}{m}^3)(5-2)\times {10}^{5}N{m}^{-2}=3kJ$

Work done by the gas the cycle $BDCB$ should be negative,

$W_2=-\text{(area of loop BDCB)}$

Now evidently triangles $ABE$ and $BCD$ are similar, the corresponding angles being equal $(i.e.,\angle EBA=\angle DBC,\angle EAB=\angle BCD)$
$\frac{(V_E-V_A)}{(P_B-P_A)}=\frac{V_C-V_D}{P_C-P_B}\Rightarrow \frac{20}{(5-2)}=\frac{(V_C-V_D)}{(6-5)}$
$\Rightarrow (V_C-V_D)=\frac{20}{3}liter$
$W_2=-\frac{1}{2}\times \frac{20}{3}\times {10}^{-3}\times 1\times {10}^5$
$=-0.33kJ$
$\therefore$ Total work done by the gas
$\Delta W=W_1+W_2=3kJ-0.33kJ=2.67kJ$
Final answer: (a)