(c) At constant volume $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow T_{2} = \left( \frac{P_{2}}{P_{1}} \right)T_{1}$ $\Rightarrow T_{2} = \left( \frac{3P}{P} \right) \times (273 + 35) = 3 \times 308 = 924K = 651℃$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.