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Two perfect gases at absolute temperatures \mathrm{T_{1}} and \mathrm{T_{2}} are mixed. There is no loss of energy. The temperature of the mixture if the masses of the molecules are \mathrm{m_{1}} and \mathrm{m_{2}} and the number of molecules in the gases are \mathrm{n_{1}} and \mathrm{n_{2}} is

Option: 1

\mathrm{\frac{\left(n_{1} T_{1}+n_{2} T_{2}\right)}{\left(n_{1}+n_{2}\right)}}


Option: 2

\mathrm{\frac{\left(\mathrm{T}_{1}+\mathrm{T}_{2}\right)}{2}}


Option: 3

\frac{\left(\mathrm{n}_{1} \mathrm{~T}_{2}+\mathrm{n}_{2} \mathrm{~T}_{1}\right)}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}


Option: 4

None of the above


Answers (1)

best_answer

According to the kinetic theory of gases, the kinetic energy of an ideal gas molecule at temperature \mathrm{T} is given by \mathrm{KE}=(3 / 2) \mathrm{kT}. And as there is no force of attraction among the molecules of a perfect gas,\mathrm{PE} of the molecule is zero. So the energy of a molecule of perfect gas,

\mathrm{\mathrm{E}=\mathrm{KE}+\mathrm{PE}=\frac{3}{2} k T+0=\frac{3}{2} k T}

Now if \mathrm{T} is the temperature of the mixture, by conservation of energy, i.e.


\mathrm{\mathrm{n}_{1} \mathrm{E}_{1}+\mathrm{n}_{2} \mathrm{E}_{2}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \mathrm{E}}
\mathrm{\text { we have } \mathrm{n}_{1}\left(\frac{3}{2} k T_{1}\right)+n_{2}\left(\frac{3}{2} k T_{2}\right)=\left(n_{1}+n_{2}\right) \frac{3}{2} k T}
\mathrm{\text { i.e. } \mathrm{T}=\frac{\left(n_{1} T_{1}+n_{2} T_{2}\right)}{\left(n_{1}+n_{2}\right)}}

Posted by

Devendra Khairwa

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