Answer
See the detailed answer below.
Work Step by Step
$\star$ We have here two processes:
$\bullet$ The first one is an isothermal process. At final of this process, the temperature is still 127$^\circ$ and the final volume is $V_2=\frac{1}{2}V_1$
$\bullet\bullet$ The second one is an isobaric process. The temperature changes but the pressure remains constant and the gas's final volume is $V_3=\frac{1}{2}V_2=\frac{1}{4}V_1$
where $V_i$ is the initial volume of the gas before starting any process.
To find the final temperature,
$\bullet$ For the first process,
$$P_1V_1=P_2V_2$$
$$P_1\color{red}{\bf\not} V_1=P_2\left(\frac{1}{2}\color{red}{\bf\not} V_1\right) $$
Hence,
$$P_2=2P_1=2(2)=\bf 4 \;\rm atm$$
$\bullet\bullet$ For the second process, where
$$P_2=P_3=\color{red}{\bf 4}\;\rm atm$$
$$\dfrac{V_2}{T_2}=\dfrac{V_3}{T_3}$$
where $T_2=T_1$, $V_2=\frac{1}{2}V_1$, and $V_3=\frac{1}{4}V_1$
$$\dfrac{\frac{1}{2}\color{red}{\bf\not} V_1}{T_1}=\dfrac{\frac{1}{4}\color{red}{\bf\not} V_1}{T_3}$$
Thus,
$$4T_3=2T_1$$
$$T_3=\dfrac{1}{2}T_1=\dfrac{1}{2}(127+273)$$
$$T_2=200\;\rm K=200-273 =\color{red}{\bf -73}^\circ\;\rm C$$
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b) We have here 3 points which are
1) The initial point at which $ (V_1,\rm2\;atm)$
2) The middle point at which $ (\frac{1}{2}V_1,\rm4\;atm)$
3) The final point at which $ (\frac{1}{4}V_1,\rm4\;atm)$
See the graph below.
