$\Rightarrow \frac{50}{100}=1-\frac{290}{\mathrm{T}_{1}} \Rightarrow \mathrm{T}_{1}=580 \mathrm{K}$
$\Rightarrow \frac{60}{100}=1-\frac{290}{\mathrm{T}_{1}^{\prime}} \Rightarrow \mathrm{T}_{1}^{\prime}=725 \mathrm{K}$
$\Rightarrow \Delta \mathrm{T}=\mathrm{T}_{1}^{\prime}-\mathrm{T}_{1}=145 \mathrm{K}$


$(A)$ the process during the path $\mathrm{A} \rightarrow \mathrm{B}$ is isothermal
$(B)$ heat flows out of the gas during the path $\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}$
$(C)$ work done during the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$ is zero
$(D)$ positive work is done by the gas in the cycle $ABCDA$
