Question

An ideal monoatomic gas undergoes a cyclic process ABCA as shown in the figure. The ratio of heat absorbed during AB to the work done on the gas during BC is –

• A. $\frac{5}{2\ln 2}$
• B. $\frac{5}{3}$
• C. $\frac{5}{4\ln 2}$
• D. $\frac{5}{6}$

Answer 1

Answer: (C)

$\frac{5}{4\ln 2}$

Answer 2

Q_AB = W_AB + DU

= nRT₀ + $\frac{nR}{\gamma - 1}$T₀ = nRT₀ $\left( 1 + \frac{1}{\gamma - 1} \right)$

= nRT₀ $\frac{\gamma}{\gamma - 1}$= nRT₀ $\frac{5/3}{\frac{5}{3} - 1}$ = $\frac{5}{2}$nRT₀

W_BC = nR2T₀ ln$\frac{1}{2}$

\ workdone on the gas

= 2 nRT₀ln2

$\frac{Q_{AB}}{W_{BC}}$ = $\frac{5}{4\mathcal{l}n2}$