Answer
a) $301\;\rm K$
b) $338\;\rm K$
Work Step by Step
a) First of all, we need to find the temperature of the gas at point 1 before it goes under the isothermal process or the other process.
We know, for an ideal gas, that
$$P_1V_1=nRT_1$$
Hence,
$$T_1=\dfrac{P_1V_1}{nR}$$
Plugging the known from the given graph;
$$T_1=\dfrac{(100\times 10^3)(2)}{(80)(8.31)}$$
$$T_1=\color{red}{\bf 301}\;\rm K$$
which is the temperature of the isothermal process.
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b) We can use the straight line equation of $y=mx+b$ where $y$ here is $P$ and $x$ here is $V$. Also, $m$ is the slope of the line, and $b$ is a constant we can find by plugging the data at point 1.
$$P=mV+b$$
The slope of the line is given by $\dfrac{P_2-P_1}{V_1-V_2}$.
$$P=\dfrac{P_2-P_1}{V_1-V_2}V+b$$
$$P=\dfrac{200,000-100,000}{1-2}V+b$$
$$P=-10^5V+b$$
Use point 1 data;
$$P_1=-10^5V_1+b$$
$$100,000=-10^5(2)+b$$
Hence,
$$b=3\times 10^5$$
Thus,
$$P=-10^5V+(3\times10^{5})$$
Therefore,
$$P=( 3-V)\times 10^5\tag 1$$
Using the ideal gas law
$$PV=nRT$$
So,
$$( 3-V)V\times 10^5=nRT$$
Hence,
$$T=\dfrac{(3V-V^2)\times 10^5}{nR}$$
Now we need to maximize $T$, so we will take the derivative of $dT/dV=0$;
$$\dfrac{dT}{dV}=0=\dfrac{(3-2V_{max})\times 10^5}{nR}$$
Hence,
$$V_{max}=1.5\;\rm m^3$$
which is the distance between the two volumes, the initial and the final.
Hence, the temperature there is given by
$$T_{max}=\dfrac{PV_{max}}{nR}=\dfrac{(150\times 10^3)(1.5)}{(80)(8.31)}$$
$$T_{\rm max}=\color{red}{\bf 338}\;\rm K$$
