Physics - Heat & Thermo Question with Solution | TestHub

From graph Process

$1\to 2$ is isobaric with $\mathrm{P}=\frac{{\mathrm{RT}}_0}{{\mathrm{V}}_0}$ Process

$2\to 3$ is isochoric with $\mathrm{V}=2{\mathrm{V}}_0$ Process

$3\to 4$ is isobaric with $\mathrm{P}=\frac{{\mathrm{RT}}_0}{2{\mathrm{V}}_0}$ Process

$4\to 1$ is isochoric with $\mathrm{V}={\mathrm{V}}_0$

Work in cycle $=\frac{{\mathrm{RT}}_0}{{\mathrm{V}}_0}.{\mathrm{V}}_0-\frac{{\mathrm{RT}}_0}{2{\mathrm{V}}_0}.{\mathrm{V}}_0=\frac{{\mathrm{RT}}_0}{2}$

${\mathrm{Q}}_{1-2}={\mathrm{nC}}_{\mathrm{P}}\Delta \mathrm{T}=\mathrm{n}.\frac{5\mathrm{R}}{2}.{\mathrm{T}}_0$

${\mathrm{Q}}_{2-3}={\mathrm{nC}}_{\mathrm{V}}\Delta \mathrm{T}=\mathrm{n}.\frac{3\mathrm{R}}{2}.{\mathrm{T}}_0$

$\therefore \left|\frac{{\mathrm{Q}}_{1-2}}{{\mathrm{Q}}_{2-3}}\right|=\frac{5}{3}$

${\mathrm{Q}}_{3-4}={\mathrm{nC}}_{\mathrm{P}}\Delta \mathrm{T}=\mathrm{n}.\frac{5\mathrm{R}}{2}.\frac{{\mathrm{T}}_0}{2}$

$\therefore \left|\frac{{\mathrm{Q}}_{1-2}}{{\mathrm{Q}}_{3-4}}\right|=2$