Three perfect gases at absolute temperatures | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The internal energy of an ideal gas is given by $U = \frac{f}{2} n R T$,where $f$ is the degrees of freedom,$n$ is the number of moles,and $T$ is

Vedclass · Accepted Answer

$\frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}$

Vedclass · Answer

(C) The internal energy of an ideal gas is given by $U = \frac{f}{2} n R T$,where $f$ is the degrees of freedom,$n$ is the number of moles,and $T$ is the absolute temperature.Since the gases are mixed and there is no loss of energy,the total internal energy of the mixture is the sum of the internal energies of the individual gases.Total internal energy $U_{total} = U_1 + U_2 + U_3$.Using $n = \frac{N}{N_A}$ (where $N$ is the number of molecules and $N_A$ is Avogadro's number),the energy is $U = \frac{f}{2} \frac{N}{N_A} R T = \frac{f}{2} N k_B T$.Assuming the gases have the same degrees of freedom $f$,the total energy is $\frac{f}{2} k_B (n_1 T_1 + n_2 T_2 + n_3 T_3) = \frac{f}{2} k_B (n_1 + n_2 + n_3) T_{mix}$.Canceling the common terms $\frac{f}{2} k_B$,we get $n_1 T_1 + n_2 T_2 + n_3 T_3 = (n_1 + n_2 + n_3) T_{mix}$.Therefore,the final temperature of the mixture is $T_{mix} = \frac{n_1 T_1 + n_2 T_2 + n_3 T_3}{n_1 + n_2 + n_3}$.

Three perfect gases at absolute temperatures $T_1, T_2$ and $T_3$ are mixed. The masses of molecules are $m_1, m_2$ and $m_3$ and the number of molecules are $n_1, n_2$ and $n_3$ respectively. Assuming no loss of energy,the final temperature of the mixture is

Similar Questions

$A$ ball is dropped from a height $h$ on a floor of coefficient of restitution $e$. The total distance covered by the ball just before the second hit is

Three charges $Q$,$+q$,and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic potential energy of the configuration is zero if $Q$ is equal to

Which of the following is not considered an organometallic compound?

$A$ heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $3: 1$. The ratio of radii of the fragments is

If the pair of straight lines given by $Ax^2 + 2Hxy + By^2 = 0$ $(H^2 > AB)$ forms an equilateral triangle with the line $ax + by + c = 0$,then $(A + 3B)(3A + B)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module