Q. Carbone monoxide is carried around a closed cycle $abc$ in which $bc$ is an isothermal process as shown in the figure. The gas absorbs $7000\ J$ of heat as its temperature increases from $300\ K$ to $1000\ K$ in going from $a$ to $b$. The quantity of heat rejected by the gas during the process $ca$ is
Explanation
(d) For path $ab:(\mathrm{\Delta}U)_{ab} = 7000\ J$
By using $\mathrm{\Delta}U = \mu C_{V}\mathrm{\Delta}T$
$7000 = \mu \times \frac{5}{2}R \times 700 \Rightarrow \mu = 0.48$
For path $ca:$
$(\mathrm{\Delta}Q)_{ca} = (\mathrm{\Delta}U)_{ca} + (\mathrm{\Delta}W)_{ca}$ …(i)
$\because(\mathrm{\Delta}U)_{ab} + (\mathrm{\Delta}U)_{bc} + (\mathrm{\Delta}U)_{ca} = 0$
$\therefore 7000 + 0 + (\mathrm{\Delta}U)_{ca} = 0 \Rightarrow (\mathrm{\Delta}U)_{ca} = - 7000\ J$ …(ii)
Also $(\mathrm{\Delta}W)_{ca} = P_{1}\left( V_{1} - V_{2} \right) = \mu R(T_{1} - T_{2})$
$= 0.48 \times 8.31 \times (300 - 1000) = - 2792.16\ J$ …(iii)
On solving equations (i), (ii) and (iii)
$(\mathrm{\Delta}Q)_{ca} = - 7000 - 2792.16 = - 9792.16\ J = - 9800\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.