Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
14 g of CO at 27° C is mixed with 16 g of O 2 at 47° C. The temperature of mixture is (vibration mode neglected)
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. $14\, g$ of $CO$ at $27^{\circ} C$ is mixed with $16\, g$ of $O _{2}$ at $47^{\circ} C$. The temperature of mixture is (vibration mode neglected)
Kinetic Theory
A
$-5^{\circ} C$
B
$32^{\circ} C$
C
$37^{\circ} C$
D
$27^{\circ} C$
Solution:
$1$ mole of $CO$ and $1$ mole of $O _{2}$ are mixed.
Net internal energy $=\frac{f_{1}}{2} R T_{ CO }+\frac{f_{2}}{2} R T_{ O _{2}}$
$=\frac{5}{2} R 300+\frac{5}{2} R 350$
$=\frac{5}{2} R(650) $
$=5 R(325) $
$=1625 \,R$
$1625=\frac{5}{2} R T_{\text {final }} \times n_{\text {final }}$
$\frac{1625 \times 2}{5}=T_{\text {final }} \times n_{\text {final }}$
$325 \times 2=T_{\text {final }} \times 2$
$T_{\text {frall }}=325 \,K$
$T_{\text {finsl }}=37^{\circ} C$