Chemistry - Thermodynamics - I Question with Solution | TestHub

In a cyclic process, internal energy change $=0$ According to $1^{\text{st }}$ law of thermodynamics $\Delta U=\Delta q+\Delta W$ As, $\Delta U=0,\Delta q=-\Delta W$ From the graph, work done $=\Delta W=$ Area under curve Here, area of circle $=\pi r^2$ $\begin{array}{rl}& =\pi {\left(\frac{25-5}{2}\right)}^2=\pi \times 10^2\\ \therefore \Delta q& =100\pi \mathrm{J}\end{array}$