🛠️ JEE🧲 Physics

Two ideal polyatomic gases at temperatures \(\boldsymbol{T}_{\mathbf{1}}\) and \(\boldsymbol{T}_{\mathbf{2}}\) are mixed…

Q1

Two ideal polyatomic gases at temperatures \(\boldsymbol{T}_{\mathbf{1}}\) and \(\boldsymbol{T}_{\mathbf{2}}\) are mixed so that there is no loss of energy. If \(\boldsymbol{F}_1\) and \(\boldsymbol{F}_2, \boldsymbol{m}_1\) and \(\boldsymbol{m}_2, \boldsymbol{n}_1\) and \(\boldsymbol{n}_2\) be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:

[JEE Main 2021, 17 Mar (Shift 1)]

a

\(\frac{{n}_{1}{T}_{1}+{n}_{2}{T}_{2}}{{n}_{1}+{n}_{2}}\)

b

\(\frac{{n}_{1}{F}_{1}{T}_{1}+{n}_{2}{F}_{2}{T}_{2}}{{n}_{1}{F}_{1}+{F}_{2}{n}_{2}}\)

c

\(\frac{{n}_{1}{F}_{1}{T}_{1}+{n}_{2}{F}_{2}{T}_{2}}{{F}_{1}+{F}_{2}}\)

d

\(\frac{{n}_{1}{F}_{1}{T}_{1}+{n}_{2}{F}_{2}{T}_{2}}{{n}_{1}+{n}_{2}}\)

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