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A thermodynamical process is shown in the figure with `p_A=3xxp_(atm)`, `V_A=2xx10^-4m^3`, `p_B=8xxp_(atm)`, `V_C=5xx10^-4m^3`. In the process AB and BC, 600J and 200J heat are added to the system. Find the change in internal energy of the system in the process CA. `[1 p_(atm)=10^5N//m^2]` A. (a) `560J`B. (b) `-560J`C. (c) `-240J`D. (d) `+240J` |
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Answer» Correct Answer - B In a cyclic process, `Q_(n et)=W_(n et)` `:.` `Q_(AB)+Q_(BC)+Q_(CA)=` area under the graph `:.` `600+200+Q_(CA)=1/2(3xx10^-4)(5xx10^5)` `=75J` `:.` `Q_(CA)=-725J=W_(CA)+DeltaU_(CA)` `=-725=` (-Area under the graph)`+DeltaU_(CA)` `=-(1/2xx11xx10^5)(3xx10^-4)+DeltaU_(CA)` Solving we get, `DeltaU_(CA)=-560J` |
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