# NEET Physics Thermodynamics Questions Solved

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 -

(1) 4200 J

(2) 5000 J

(3) 9000 J

(4) 9800 J

(4) For path ab : ${\left(\Delta U\right)}_{ab}=7000\text{\hspace{0.17em}}J$

By using $\Delta U=\mu {C}_{V}\Delta T$

$7000=\mu ×\frac{5}{2}R×700⇒\mu =0.48$

For path ca :

${\left(\Delta Q\right)}_{ca}={\left(\Delta U\right)}_{ca}+{\left(\Delta W\right)}_{ca}$ ….(i)

${\left(\Delta U\right)}_{ab}+{\left(\Delta U\right)}_{bc}+{\left(\Delta U\right)}_{ca}=0$

$7000+0+{\left(\Delta U\right)}_{ca}=0⇒{\left(\Delta U\right)}_{ca}=-7000\text{\hspace{0.17em}}J$ ….(ii)

Also ${\left(\Delta W\right)}_{ca}={P}_{1}\left({V}_{1}-{V}_{2}\right)=\mu R\left({T}_{1}-{T}_{2}\right)$

$=0.48×8.31×\left(300-1000\right)=-2792.16\text{\hspace{0.17em}}J$ ….(iii)

on solving equations (i), (ii) and (iii)

${\left(\Delta Q\right)}_{ca}=-7000-2792.16=-9792.16\text{\hspace{0.17em}}J$ = –9800 J

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