Two perfect gases at absolute temperatures $T_{1} a n d T_{2}$ are mixed. There is no loss of energy in this process. If $n_{1} a n d n_{2}$ are the respective number of molecules of the gases, the temperature of the mixture will be

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Average kinetic energy per molecule of a perfect gas $= \frac{3}{2} k T$
$∴$ Average KE of molecules of the first gas $\frac{3}{2} n_{1} k T_{1}$
Average KE of molecules of the second gas = $\frac{3}{2} n_{2} k T_{2}$
$∴$ Total KE of the molecules of the two gases before they are mixed is
$K = \frac{3}{2} n_{1} k T_{1} + \frac{3}{2} n_{2} k T_{2} = \frac{3}{2} (n_{1} T_{1} + n_{2} T_{2}) k (1)$
If T is the temperature of the mixture, the kinetic energy of the molecules $(n_{1} + n_{2})$ in the mixture is
$K^{'} = \frac{3}{2} (n_{1} + n_{2}) k T (2)$
Since there is no loss of energy K=K'. Equating (1) and (2) we get
$n_{1} T_{1} + n_{2} T_{2} = (n_{1} + n_{2}) T o r T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Suggest Corrections

Similar questions

Two ideal gases at absolute temperatures $T_{1}$ and $T_{2}$ are mixed. There is no loss of energy in this process if $n_{1}$ and $n_{2}$ are the respective number of molecules of the gases, the temperature of the mixture will be

Q. Two Perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m1 and m2 and the number of molecules in the gases are

n_{1}

and

n_{2}

$n_{1}$ and $n_{2}$ moles of two ideal gases (mol. wt. $M_{1}$ and $M_{2}$ ) respectively at temperature $T_{1}$ K and $T_{2}$ K are mixed. Assuming that no loss of energy, the temperature of mixture becomes

Two perfect gases at absolute temperatures $T_{1} a n d T_{2}$ are mixed. There is no loss of energy in this process. If $n_{1} a n d n_{2}$ are the respective number of molecules of the gases, the temperature of the mixture will be

Join BYJU'S Learning Program

Two perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy in this process. If n1 and n2 are the respective number of molecules of the gases, the temperature of the mixture will be

Two perfect gases at absolute temperatures $T_{1} a n d T_{2}$ are mixed. There is no loss of energy in this process. If $n_{1} a n d n_{2}$ are the respective number of molecules of the gases, the temperature of the mixture will be