Question

A monoatomic gas undergoes a cycle consisting of two iso-thermal and two isobaric processes. IF minimum and maximum temperatures of the gas during the cycle are $T_1=400K$ and $T_2=800K$ and ratio of maximum to minimum volume is $r=4$ then the efficiency of working cycle is -

• A. $\frac{2\log ,2}{5+4{\log }_{e}2}$
• B. $\frac{2\log ,2}{5+2{\log }_{e}2}$
• C. $\frac{4\log ,2}{5+4{\log }_{e}2}$
• D. $\frac{2\log ,2}{5+8{\log }_{e}2}$

Answer 1

Answer: (A)

$\frac{2\log ,2}{5+4{\log }_{e}2}$

$\text{Given:-}$ Minimum temperatures $T_1=400K$

Maximum temperatures $T_2=800K$

$\text{Expalantion:-}$

Process CB is done at minimum temperature, while process DA is done at maximum temperature.

Process

$A\to B\left(isobaric\right)$

$\frac{V_A}{V_B}=\frac{T_A}{T_B}=\frac{800}{400}=2$

Hence , $V_B=2V_0$

Process

$B\to C\left(isothermal\right)C\to D\left(isobaric\right)$

$\frac{V_D}{V_C}=\frac{T_D}{T_C}=V_D=2V_0$

$D\to A\left(isothermal\right)$

$W_{AB}=P(V_B-V_A)=nRDT$

$=nR(-400)=-400nR$

$W_{BC}=nRTln\frac{V_C}{V_B}=nR400ln\frac{V_0}{2V_0}$

$=-400nRln2$

$W_{CD}=P(V_D-V_C)=nR\left(400\right)=400nR$

$W_{DA}=nRTln\frac{V_A}{V_D}$

$=nR\times 800\times \times ln\frac{2V_0}{V_0}=800nRln2$

$\Delta W=400nRln2\Delta Qheatisextracted{Q}_{CD}=n{C}_P\Delta T=n\frac{5}{2}R400=1000nR$

$=800nRln2=1000nR+800nRln2$

$Efficiency=\frac{\Delta W}{\Delta Q}$

$\eta =\frac{2ln2}{5+4ln2}\times 100$

$\text{Hence the correct option is A}$