Question 85: Thermodynamics & KTG

Question

Carnot cycle of an engine is given below
Total work done by the gas in one cycle is
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Accepted Answer

Total work done by the gas
${W_{Total}} = {W_{P 	o Q}} + {W_{Q 	o R}} + {W_{R 	o S}} + {W_{S 	o P}}$
Work done in isothermal process by ideal gas
${W_{P 	o Q}} =  \mu RT_1\log {{{V_2}} \over {{V_1}}}$ ..... (i)
Work done in adiabatic process by ideal gas
${W_{Q 	o R}} = {{ \mu R({T_1} - {T_2})} \over {r - 1}}$ ..... (ii)
Work done in isothermal process by ideal gas,
${W_{R 	o S}} =  -  \mu R{T_2}\log {{{V_3}} \over {{V_4}}}$ ...... (iii)
Work done in adiabatic process, by ideal gas
${W_{S 	o P}} = {{ -  \mu R({T_1} - {T_2})} \over {r - 1}}$ ..... (iv)
Adding Eqs. (i), (ii), (iii) and (iv) we get
${W_{Total}} = \mu R{T_1}\log {{{V_2}} \over {{V_1}}} - \mu R{T_2}\log {{{V_3}} \over {{V_4}}}$

Answer

$\mu R{T_2}\log {{{V_2}} \over {{V_1}}} - \mu R{T_1}\log {{{V_3}} \over {{V_4}}}$

Answer

$\mu R{T_1}\log {{{V_2}} \over {{V_1}}} - \mu R{T_2}\log {{{V_3}} \over {{V_4}}}$

Answer

$\mu R{T_1}\log {{{V_2}} \over {{V_1}}} + \mu R{T_2}\log {{{V_3}} \over {{V_4}}}$

Answer

Zero

Thermodynamics & KTG

Carnot cycle of an engine is given below Total work done by the gas in one cycle is

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