Q. 10 moles of an ideal monoatomic gas at $10℃$ is mixed with 20 moles of another monoatomic gas at $20℃.$ Then the temperature of the mixture is
A) $15.5℃$
B) $15℃$
C) $16℃$
D) $16.6℃$
Explanation
(d) $n_{1}C_{v}\mathrm{\Delta}T_{1} = n_{2}C_{v}\mathrm{\Delta}T_{2}$
$10 \times (T - 10) = 20(20 - T)$
$T - 10 = 40 - 2T$
$3T = 50 \Rightarrow T = 16.6℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.