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InterviewSolution
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Five moles of an ideal gas are taken in a carnot engine working between `100^(@)C and 30^(@)C`. The useful work done in 1 cycle is `420J`. Calculate ratio of volume of the gas at the end and at the beginning of the isothermal expansion. Take `R=8.4J mol e^(-1)K^(-1)` |
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Answer» Correct Answer - `1.153` Here, `n=5, T_(1)= 100^(@)C= (100+273)K = 373K`, `T_(2)= 30^(@)C=(30+273)K=303K`, `W=420J, (V_(2))/(V_(1))=?, R= 8.4J mol e^(-1)K^(-1)` As `(Q_(1))/(Q_(2))=(T_(1))/(T_(2))=(373)/(303) :. Q_(1)(373)/(303)Q_(2)`....(i) Now, `W=Q_(1)-Q_(2)=(373)/(303)Q_(2)-Q_(2)=(70)/(303)Q_(2)` `:. Q_(2)=(373)/(70)W= (303)/(70)xx420=1818J` From (i), `Q_(1)=(373)/(303)xx1818= 2238J` Work done during isothermal expansion `W_(1)=Q_(1)= 2.3026 n RT_(1)"log"_(10)(V_(2))/(V_(1))` `2238= 2.3026xx5xx8.4xx373 "log"_(10)(V_(2))/(V_(1))` `"log"_(10)(V_(2))/(V_(1))=(2238)/(2.3026xx5xx8.4xx373)=0.0620` `(V_(2))/(V_(1))= antilog0.0620= 1.153` |
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