One mole of ideal monatomic gas undergoes the cycleshown in Fig. 3, consisting o

Question

One mole of ideal monatomic gas undergoes the cycleshown in Fig. 3, consisting of an isothermal process AB at temperature 400 K, an isobaric process BC,and an isometric process CA (between pressures of 100 kPa and 200 kPa). The work done by the gas during the isothermal process is 2305 J.
(a) Find the volumes VA = VC and VB, and the temperature TC.
(b) Find the work done for the entire cycle
(c) Find the amounts of heat for each of the three processes.
(d) Find the efficiency of this heat engine.

Explanation / Answer

No.of moles n = 1 temperature at A is T = 400 K pressure at A is P = 200 kPa = 2*10^5Pa ffrom the relation PV =nRT volume at A is V = nRT / P    where R = gas constant = 8.314 J / mol K                           = 0.016628 m^ 3 The work done by the gas during the isothermal process is W = 2305 J we know W = nRT ln(V ' / V )   where V ' = volume at point B     ln( V ' / V ) = W / nRT                       = 0.693          V ' / V = e^0.693                     = 2               V ' = 2V = 0.033256 m^ 3 along path BC pressure is constant. So, V " / V ' = T" / T '    where T ' = T = 400 K            V" = V            V' = 2V So, T " = T'( V" / V ')             = 400 ( V /2V )             = 200 K (b). work done in BC is W ' = P ( V" - V ')                                          = 100kPa ( V -2V ) = -PV                                          = -100*10^3 Pa *0.016628 m^ 3                                          = -1662.8 J work done in CA is W " = 0   SInce volume is constant net work done w = W + W ' +W"                             = 642.2 J (c). heat in AB is Q = nRT ln( V ' / V )                                = W                                = 2305 J heat in BC is Q ' = n*Cp*(T"-T')   where Cp = specific heat at constant pressure = 2.5 R                            = 1*2.5*8.314 * ( 200-400)                            = -4157 J heat in CA is Q " = n*Cv*(T-T") where Cv = 1.5 R                            = 1* 1.5*8.314 *(400-200)                            = 2494.2 J (d). efficiency n = net work / heat taken                          = w / ( Q + Q " )                          = 0.1338                          = 13.38 %

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.