Question 81: Thermodynamics & KTG

Question

Carnot cycle of an engine is given below

Total work done by the gas in one cycle is

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

Choose the correct answer: