Question

$P-T$ diagram of one mole of an ideal monoatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is

• A. $2.5RT$
• B. $-2RT$
• C. $1.5RT$
• D. $-3.5RT$

Answer 1

Answer: (A)

$2.5RT$

let $A=1,B=2,C=3,D=4$
now $P_1=P_4=P$ & $P_2=P_3=2P$

so by ideal gas equation we have
$V_1$=$\frac{3RT}{P}$ $V_2$=$\frac{RT}{2P}$

now,
1) for process 1-2
$W_{1-2}$=$\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$
so $W_{1-2}=\frac{2P\frac{RT}{2P}-P\frac{3RT}{P}}{0.4}=-5RT$

2)for process 3-4
$W_{3-4}=\frac{2P\frac{2RT}{P}-P\frac{5RT}{P}}{0.4}=-2.5RT$

3) for process 4-1
$W_{4-1}=P_1(V_4-V_1)$
$W_{4-1}=P[\frac{5RT}{P}-\frac{3RT}{P}]=2RT$

4) for process 2-3
$W_{2-3}=2P[\frac{2RT}{P}-\frac{RT}{2P}]=3RT$

now adding[1,2,3,4] we get the net work done
$W_{total}$=$W_{1-2}+W_{2-3}+W_{3-4}+W_{4-1}$
=$-5RT+3RT+-2.5RT+2RT=-2.5RT$