Question

An ideal gas is taken through a cyclic thermodynamic process through four steps $1,2,3,4$ respectively. The amounts of heat involved in these steps are $Q_1=5960\text{J},Q_2=5585\text{J},Q_3=-2980\text{J}$ and $Q_4=3645\text{J}$ respectively. The corresponding quantities of work involved are $W_1=2200\text{J},W_2=8250\text{J},W_3=-1100\text{J}$ and $W_4$ respectively. The value of $W_4$ is

• A. $1865\text{J}$
• B. $2860\text{J}$
• C. $6620\text{J}$
• D. $8620\text{J}$

Answer 1

The correct option is B $2860\text{J}$

In a cyclic process , $\Delta U=0$
$\Rightarrow$From, first law of thermodynamics
$Q=\Delta U+W...\left(i\right)$
For the cyclic process $Q=Q_1+Q_2+Q_3+Q_4$
Net work done in cyclic process is,
$W=W_1+W_2+W_3+W_4$
Hence from Eq$\left(i\right)$,
$Q_1+Q_2+Q_3+Q_4=W_1+W_2+W_3+W_4$
$\Rightarrow W_4=(Q_1+Q_2+Q_3+Q_4)-(W_1+W_2+W_3)$
$W_4=(5960+5585-2980+3645)-(2200+8250-1100)$
$\therefore W_4=2860\text{J}$