Question

Three perfect gases at absolute temperatures T1,T2 and T3 are mixed.If number of molecules of the gases are n1,n2 and n3 respectively then temperature of mixture will be (assume no loss of energy)?

Answer 1

Answer on Question #42746, Physics, Molecular Physics | Thermodynamics

Three perfect gases at absolute temperatures $T_{1}, T_{2}$ and $T_{3}$ are mixed. If numbers of molecules of the gases are $n_{1}, n_{2}$ and $n_{3}$ respectively then temperature of mixture will be (assume no loss of energy)?

Solution:

An ideal gas can be characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them may be deduced from kinetic theory and is called the ideal gas law:

$$PV = nkT$$

where

- $n$ = number of molecules

- $k$ = Boltzmann constant $= 1.38066 \times 10^{-23} \, \text{J/K} = 8.617385 \times 10^{-5} \, \text{eV/K}$

From data we have

$$n_{1} + n_{2} + n_{3} = n$$

By energy conservation

$$P_{1}V_{1} + P_{2}V_{2} + P_{3}V_{3} = PV$$

Thus,

$$n_{1}kT_{1} + n_{2}kT_{2} + n_{3}kT_{3} = nkT$$

So, the final temperature is

$$T = \frac{n_{1}T_{1} + n_{2}T_{2} + n_{3}T_{3}}{n_{1} + n_{2} + n_{3}}$$

Answer. $T = \frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}$

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