| \[{{\operatorname{Q}}_{1}}=6000J, {{Q}_{2}}=-5500J\] |
| \[{{\operatorname{Q}}_{3}} =-3000J, {{Q}_{4}}= 3500J\] |
| \[{{\operatorname{W}}_{1}} = 2500J, {{W}_{2}} = -1000J\] |
| \[{{\operatorname{W}}_{3}}=1200J,{{W}_{4}}=xJ\] |
| The ratio of net work done by the gas to the total heat absorbed by the gas is n. The value of x and n are nearly. |
A) 500, 7.5%
B) 700, 10.5%
C) 1000, 21%
D) 1500,15%
Correct Answer: B
Solution :
\[\operatorname{In} a cyclic process \Delta U=0\] \[{{1}^{st}}\]law of thermodynamic \[\operatorname{W}=\Delta U+Q\] \[{{\operatorname{W}}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}=\Delta U+{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] \[2500+\left( -1000 \right)+\left( -1200 \right)+\operatorname{x}=0+6000-5500\]\[300+x=1000\] \[\operatorname{x}=700J\] Efficiency \[\eta =\frac{\Delta \operatorname{W}}{{{\operatorname{Q}}_{1}}+{{Q}_{4}}}=\frac{\Delta \operatorname{Q}}{{{\operatorname{Q}}_{1}}+{{Q}_{4}}}=\frac{1000}{9500}\times 100=10.5%\]
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