(c) $C_{p} = 8$ cal $(mol℃)^{- 1},\ C_{v} = C_{p} - R = 8 - 2 = 6$ cal $(mol\ ℃)^{- 1}$ $dU = mC_{v}\left( T_{2} - T_{1} \right) = 5 \times 6(20 - 10) = 300\ cal$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.