Question

At what temperature will the mean molecular energy of a perfect gas be one-third of its value of 27${}^o$C?

• A. 10${}^o$C
• B. 10${}^1$K
• C. 10${}^2$K
• D. 10${}^3$J

Answer 1

Answer: (C)

10${}^2$K

Mean $KE\propto T$
therefore at T=27 which in kelvin becomes 300 K
So KE will be 1/3 when temperature becomes 1/3 which is 300/3=100K
option (C)