Question

Determine the work done by an ideal gas undergoing a cyclic process from $1→4→3→2→1$ . Given $P_1=10^{5}Pa,P_0=3×10^{5}Pa,P_3=4×10^{5}Pa$ and $V_2−V_1=10L$ .

• A. $740J$
• B. $750J$
• C. $730J$
• D. $745J$

Answer 1

Answer: (B)

From figure
$\frac{{\text{V}}_4−{\text{V}}_3}{{\text{V}}_2−{\text{V}}_1}=\frac{{\text{P}}_3−{\text{P}}_0}{{\text{P}}_0−{\text{P}}_1}⇒\frac{{\text{V}}_4−{\text{V}}_3}{10}=\frac{4×10^5−3×10^5}{3×10^5−10^5}$
$V_4−V_3=5L$
Now, work done
$\text{W}=\left(\frac{1}{2}×1×2×10^5−\frac{1}{2}×5×1×10^5\right)×10^{−3}=750J$