Question

One mole of an ideal monatomic gas is taken along the cycle abcda

Answer 1

your question is incomplete. A complete question is -----> One mole of an ideal monatomic gas is taken along the cycle ABCDA where AB is isochoric, BC is isobaric, CD is adiabatic and DA is isothermal.  Find the efficiency of the cycle.  It is given that TC/ TA=4/1 , VA/ VD=1/16 .


solve :- given, $\frac{V_A}{V_D}=\frac{1}{16}$
if we assume, $V_A$= v
then, volume , $V_D$ = 16v

similarly, we can assume temperature, $T_C=T$
then, temperature , $T_A$ = 4T

but path DA is isothermal..
so, $T_A=T_D=T$

as path CD is adiabatic . so, $T_CV_C^{\gamma-1}=T_DV_C^{\gamma-1}$

or, $(4T)V_C^{\gamma-1}=T(16v)^{\gamma-1}$
as gas is monoatomic so, $\gamma=\frac{5}{3}$

so, $V_C^{5/3-1}=\frac{(16v)^{5/3-1}}{4}$

hence, $V_C=2v$


as path AB is isochoric process.
so, $V_A=V_B=v$

Path BC is isobaric process
$\frac{V_B}{T_B}=\frac{V_C}{T_C}$

or, $\frac{v}{T_B}=\frac{2v}{T}$

or, $T_B=0.5T$

now, heat absorbed in path AB = Cv (Ta - Tb)
= 3R/2 (T - 0.5T) = 3RT/4

heat absorbed in path BC = cp(TC - Ta)
= 5R/2 (4T - 2T) = 5RT

heat rejected in path DA = -workdone by the gas = nRTlnVA/VD = 1 × R × Tln(v/16v) = RTln16

so, efficiency of cycle = 1 - RTln16/(5.75RT)

= 1 - ln16/5.75

= 0.517