Q. 22 g of $CO_{2}\ at\ 27℃$ is mixed with 16 g of oxygen at 37$℃$. The temperature of the mixture is
A) 32$℃$
B) 27$℃$
C) 37$℃$
D) 30$℃$
Explanation
(a) $22\ g\ of\ CO_{2}$ is half mole of $CO_{2}\ ie,\ n_{1} = 0.5$
16 g of $O_{2}$ is half mole of $O_{2}\ ie,\ n_{2} = 0.5\ $
$\therefore\ T = \ \frac{n_{1}T_{1} + n_{2}T_{2}}{n_{1} + n_{2}}$
= $\frac{0.5 \times (27 + 273) + 0.5(37 + 273)}{0.5 + 0.5}$
= 305 K
= 305$- 273 = 32℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.