MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heat Capacity

Two ideal pol...

Question

Two ideal polyatomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed so that there is no loss of energy. If $F_{1}$ and $F_{2}$ , $m_{1}$ and $m_{2}$ , $n_{1}$ and $n_{2}$ be the degrees of freedom, masses, the number of molecules of the first and second gas respectively, the temperature of the mixture of these two gases is:

A

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

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

B

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

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

C

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

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

D

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

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

Open in App

Solution

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

Step 1: Given
Change in internal energy: $∆ U = 0$ (since there in no loss of energy)
Temperature of first gas is $T_{1}$
Temperature of second gas is $T_{2}$
Degree of freedom of first gas is $F_{1}$
Degree of freedom of second gas is $F_{2}$
Mass of first gas is $m_{1}$
Mass of second gas is $m_{2}$
Number of molecules of first gas is $n_{1}$
Number of molecules of second gas is $n_{2}$
Assume the final temperature of the mixture to be $T$
Step 2: Formula Used
Internal energy of a gas is given by
$U = \frac{n F k T}{2}$
Where, $n$ is the number of moles, $F$ is the degree of freedom, $k$ is the Boltzmann's constant and $T$ is the temperature.
Step 3: Find the temperature of the mixture
Since there is no loss of internal energy, sum of initial internal energies will be equal to final internal energies
${U_{1}}_{i n i t i a l} + {U_{2}}_{i n i t i a l} = {U_{1}}_{f i n a l} + {U_{2}}_{f i n a l} \Rightarrow \frac{n_{1} F_{1} k T_{1}}{2} + \frac{n_{2} F_{2} k T_{2}}{2} = \frac{n_{1} F_{1} k T}{2} + \frac{n_{2} F_{2} k T}{2} \Rightarrow n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2} = n_{1} F_{1} T + n_{2} F_{2} T$
Solve the above equation for the final temperature
$n_{1} F_{1} T + n_{2} F_{2} T = n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2} \Rightarrow T (n_{1} F_{1} + n_{2} F_{2}) = n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2} \Rightarrow T = \frac{n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2}}{n_{1} F_{1} + n_{2} F_{2}}$
Hence, the correct option is option A.

Suggest Corrections

thumbs-up

33

Similar questions

Q. Two monoatomic ideal gas at temperature

T_{1}

T_{2}

are mixed. There is no loss of energy. If the masses of molecules of the two gases are

m_{1}

m_{2}

and number of their molecules are

n_{1}

n_{2}

respectively. The temperature of the mixture will be:

Q. Two perfect gases at absolute temperatures

T_{1}

T_{2}

are mixed. There is no loss of energy. The temperature of mixture, if masses of molecules are

m_{1}

m_{2}

and the number of molecules in the gases are

n_{1}

n_{2}

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

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

Q. Three perfect gases at absolute temperatures

T_{1}, T_{2}

T_{3}

are mixed. The masses of molecules are

m_{1}, m_{2}

m_{3}

and the number of molecules are

n_{1}, n_{2}

n_{3}

respectively. Assuming no loss of energy, the final temperature of the mixture is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Thermodynamic Processes

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XIII Physics

Join BYJU'S Learning Program

CrossIcon