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Carbon 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 ..... $J$
$4200 $
$5000 $
$9000 $
$9800 $
Solution
(d) For path $ab : {(\Delta U)_{ab}} = 7000\;J$
By using $\Delta U = \mu {C_V}\Delta T$
$7000 = \mu \times \frac{5}{2}R \times 700 \Rightarrow \mu = 0.48$
For path $ca : {(\Delta Q)_{ca}} = {(\Delta U)_{ca}} + {(\Delta W)_{ca}}$ ….$(i)$
${(\Delta U)_{ab}} + {(\Delta U)_{bc}} + {(\Delta U)_{ca}} = 0$
$\therefore $$7000 + 0 + {(\Delta U)_{ca}} = 0 \Rightarrow {(\Delta U)_{ca}} = – 7000\;J$ …$(ii)$
Also ${(\Delta W)_{ca}} = {P_1}({V_1} – {V_2}) = \mu R({T_1} – {T_2})$
$ = 0.48 \times 8.31 \times (300 – 1000) = – 2792.16\;J$ ….$(iii)$
on solving equations $(i), (ii)$ and $ (iii)$
${(\Delta Q)_{ca}} = – 7000 – 2792.16 = – 9792.16\;J$$ = – 9800\;J$
