$$(V_1, P_1)=(15\ lit, 2\ atm)$$
$$(V_2, P_2)=(4\ lit, 10\ atm)$$
Eq. of line
$$P-P_1=\dfrac{P_1-P_2}{V_1-V_2}(V-V_1)$$
$$P=P_1+\dfrac{P_1-P_2}{V_1-V_2}(V-V_1)$$
$$\dfrac{nRT}{V}=P_1+\left( \dfrac{P_1-P_2}{V_1-V_2}\right)V-\left( \dfrac{P_1-P_2}{V_1-V_2}\right)V_1$$
$$T=\dfrac{1}{nR}\left\{ P_1V +\left( \dfrac{P_1-P_2}{V_1-V_2}\right)V^2-\left( \dfrac{P_1-P_2}{V_1-V_2}\right) V_1.V\right\}$$
For $$T$$ to be max $$\dfrac{dT}{dV}=0$$
$$\dfrac{dT}{dV}=\dfrac{1}{nR}\left\{ P_1+2\left( \dfrac{P_1-P_2}{V_1-V_2}\right)V-\left( \dfrac{P_1-P_2}{V_1-V_2}\right)V_1\right\}=0$$
$$2\left( \dfrac{P_1-P_2}{V_1-V_2}\right)V=\dfrac{P_1V_1 -P_2V_1}{V_1-V_2}-P_1=\dfrac{P_1V_1 -P_2V_1 -P_1V_1+P_1V_2}{V_1 -V_2}=\dfrac{P_1V_2-P_2V_1}{(V_1-V_2)}$$
$$V=\dfrac{P_1V_2-P_2V_1}{2(P_1-P_2)}=\dfrac{2\times 4-10\times 15}{2(2-10)}=8.875$$
$$T_{max}$$, for $$1\ mole$$, $$n=1$$
$$T_{max}=\dfrac{1}{R}\left\{ 2\times 8.875+\left( \dfrac{2-10}{15-4}\right)\times (8.875)^2-\left( \dfrac{2-10}{15-4}\right)\times 15\times 8.875 \right\}$$
$$=\dfrac{1}{R}\left\{ 17.75+\left( \dfrac{-8}{11}\right)\times 78.7656-\left( \dfrac{-8}{11}\right) \times 133.125 \right\}$$
$$=\dfrac{1}{R}\left\{ 17.75+96.82+(-57.28)\right\}=\dfrac{1}{R}\times 57.286=\dfrac{57.29}{0.0821}=697.80\ K$$