Chapter 10 - Temperature and Kinetic Theory - Learning Path Questions and Exercises - Multiple Choice Questions - Page 380: 14

(b) $\frac{5}{2}nRT$

Since at room temperature, vibrational energy is not active, only translation and rotation energy will contribute to internal energy. So, Internal energy is addition of $\frac{3}{2}nRT$ (translation energy) and $1nRT$ (rotational energy), which makes it $\frac{5}{2}nRT$.