Consider one mole of helium gas enclosed in a container at initial pressure $$P1$$ and volume $$V1$$. It expands isothermally to volume $$4V1$$. After this, the gas expands adiabatically and its volume becomes $$32V1$$. The work done by the gas during isothermal and adiabatic expansion processes are W𝑖𝑠𝑜 and W𝑎𝑑𝑖𝑎, respectively. If the ratio Wiso Wadia $$=fln$$⁡$$2$$, then 𝑓 is $$______$$

Question

Consider one mole of helium gas enclosed in a container at initial pressure $$P1$$ and volume $$V1$$. It expands isothermally to volume $$4V1$$. After this, the gas expands adiabatically and its volume becomes $$32V1$$. The work done by the gas during isothermal and adiabatic expansion processes are W𝑖𝑠𝑜 and W𝑎𝑑𝑖𝑎, respectively. If the ratio Wiso Wadia $$=fln$$⁡$$2$$, then 𝑓 is $$______$$

Infinity Learn · Accepted Answer

For Adiabatic process, $$P144V15/3=P232V15/3$$⇒$$P2=P14185/3=P1128Wadi$$ $$=P1V1$$−$$P2V2$$γ−$$1=P1V1$$−$$P112832V153$$−$$1$$ $$=P1V1(3/4)2/3=98P1V1Wiso$$ $$=P1V1In$$⁡$$4V1V1=2P1V1ln$$⁡$$2Wiso$$ Wadio $$=2P1V1ln$$⁡$$298P1V1=169ln$$⁡$$2=fln$$⁡$$2here$$ , $$f=169=1.7778$$≈$$1.78$$

Thermodynamic processes

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Thermodynamic processes

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For Adiabatic process, P144V15/3=P232V15/3⇒P2=P14185/3=P1128Wadi =P1V1−P2V2γ−1=P1V1−P112832V153−1 =P1V1(3/4)2/3=98P1V1Wiso =P1V1In⁡4V1V1=2P1V1ln⁡2Wiso Wadio =2P1V1ln⁡298P1V1=169ln⁡2=fln⁡2here , f=169=1.7778≈1.78

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