Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat exchanged in the steps are $Q_1=+5960\text{J},Q_2=-5600\text{J},Q_3=-3000\text{J},Q_4=+3600\text{J}$ respectively. The corresponding quantities of internal energy changes are $\Delta U_1=+3760\text{J},\Delta U_2=-4800\text{J},\Delta U_3=-1800\text{J}$ and $\Delta U_4$. Find the value of $\Delta U_4$ and work done during the cyclic process.

• A. $+2840\text{J},+960\text{J}$
• B. $+2830\text{J},+900\text{J}$
• C. $-2930\text{J},-960\text{J}$
• D. $-2930\text{J},+960\text{J}$

Answer 1

The correct option is A $+2840\text{J},+960\text{J}$

In a cyclic process, total change in internal energy is zero.
$\Delta U=[\Delta U_1+\Delta U_2+\Delta U_3+\Delta U_4]=0$
From the data given in the question,
$+3760-4800-1800+\Delta U_4=0$
$\Rightarrow \Delta U_4=+2840\text{J}$
Using first law of thermodynamics for cyclic process, we can say that
$\Delta W=\Delta Q$
$=+5960-5600-3000+3600=+960\text{J}$.
Hence, option (a) is the correct answer.