Question

Refer to the Carnot cycle of an ideal gas shown in the figure Let $W_{a\to b},W_{b\to c},W_{c\to d}$ and $W_{d\to a}$ represent the work done by the system during the processes $a\to b,b\to c,c\to d$ and $d\to a$ respectively. Consider the following relations
(1) $W_{a\to b}+W_{b\to c}+W_{c\to d}+W_{d\to a}>0$
(2) $W_{a\to b}+W_{b\to c}+W_{c\to d}+W_{d\to a}<0$
(3) $W_{a\to b}+W_{c\to d}>0$
(4) $W_{b\to c}+W_{d\to a}=0$
Which of the above relation is/are true

• A. $1$ and $3$ only
• B. $3$ and $4$ only
• C. $1,3$ and $4$ only
• D. $2,3$ and $4$ only

Answer 1

The correct option is D $2,3$ and $4$ only

$\text{Solution}:$

$W_{a-b}$ is isothermal, $W_{b-c}$ is adiabatic, $W_{c-d}$ is isothermal (Negative), $W_{d-a}$ is adiabatic (Negative).

Net work done during the complete cycle

$W=W_{a-b}+W_{b-c}+(-W_{c-d})+(-W_{d-a})$

$W=W_{a-b}-W_{c-d}$

$[As{W}_{b-c}=W_{d-a}]$

i.e., $W_{a-b}+W_{c-d}>0$ and $W_{a-b}+W_{b-c}+W_{c-d}+W_{d-a}>0$

$\because W_{b-c}=W_{d-a}$

$\therefore W_{b-c}+(-)W_{d-a}=0$

$\text{Hence D is the correct option}$