Answer
$$1.98$$
Work Step by Step
The process $2 \rightarrow 3$ is adiabatic, so $$T_{2} V_{2}^{\gamma-1}=T_{3} V_{3}^{\gamma-1} .$$ Using the result from part $(\mathrm{a}),$ $$V_{3}= 4.00 V_{1}, V_{2}=V_{1},$$ and $$\gamma=1.30$$ we obtain
$$
\frac{T_{3}}{T_{1}}=\frac{T_{3}}{T_{2} / 3.00}=3.00\left(\frac{V_{2}}{V_{3}}\right)^{\gamma-1}=3.00\left(\frac{1}{4.00}\right)^{0.30}=1.98
$$
