JEE Main & Advanced Physics Thermodynamical Processes Question Bank Graphical Questions

  • question_answer
    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 [SCRA 1994]

    A)            4200 J                                      

    B)            5000 J

    C)            9000 J                                      

    D)            9800 J

    Correct Answer: D

    Solution :

                       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) \[\because \]\[{{(\Delta U)}_{ab}}+{{(\Delta U)}_{bc}}+{{(\Delta U)}_{ca}}=0\] \[\therefore \]\[7000+0+{{(\Delta U)}_{ca}}=0\Rightarrow {{(\Delta U)}_{ca}}=-7000\ J\]              ?.(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) \[{{(\Delta Q)}_{ca}}=-7000-2792.16=-9792.16\ J\]\[=-9800\ J\]


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