First slide

Law of equipartition of energy

Question

Two perfect gases at absolute temperature $T_{1} a n d T_{2}$ are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m $m_{1} a n d m_{2} a n d t h e n u m b e r o f m o l e c u l e s i n t h e g a s e s n_{1} a n d n_{2}$ . Then final temperature of the mixture is

Moderate

A

$\frac{n_{1} T_{1} + n_{2} T_{2}}{2}$
B

$\frac{n_{1} T_{2} + n_{2} T_{1}}{n_{1} + n_{2}}$
C

$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$
D

$\frac{T_{1} + T_{2}}{2}$

Solution

energy of each molecule at temperature T is $E = \frac{3}{2} k T$
as energy is conserved $n_{1} E_{1} + n_{2} E_{2} = (n_{1} + n_{2}) E$
$n_{1} \frac{3}{2} k T_{1} + n_{2} \frac{3}{2} k T_{2} = (n_{1} + n_{2}) (\frac{3}{2}) k T$
$T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App