Question

Carbon monoxide is carried around a closed cycle $abc$ in which $bc$ is an isothermal process as shown in the figure. The gas absorbs $7000J$ of heat as its tempertur increases from $300K$ to $1000K$ in going from $a$ to $b$. The quantity of heat rejected by the gas during the process $ca$ is

• A. $4200J$
• B. $5000J$
• C. $9000J$
• D. $9800J$

Answer 1

The correct option is D $9800J$

$\text{Given}:$ $\Delta Q_{ab}=+7000$ , $T_a=300K,T_b=1000K$

$\text{Solution}:$

$\Delta Q_{ab}=+7000=\mu C_v(1000-300)$

For the process ca: $T_a=300K,T_c=T_b=1000K$

$(\Delta Q{)}_{ca}=\mu C_P(300-1000)=-\mu C_P\times 700=-\mu (C_V+R)700$

For carbon monoxide : $T_a=300K,T_c=T_v=1000K$

$(\Delta Q{)}_{ca}=\mu C_P(300-1000)=-\mu C_P\times 700=-\mu (C_V+R)700$

For carbon monoxide : $\gamma =\frac{7}{5}{C}_V=\frac{R}{\gamma -1}=\frac{R}{\frac{7}{5-1}}=\frac{5R}{2}$

Hence, form Eq. (i)

$7000=\mu \frac{5R}{2}\times 700\mu R=\frac{20}{5}=4$

$\Delta Q_{ca}=-(700+4\times 700)=-9800J$

Negative sign shows that heat is ejected.

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