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Two perfect monoatomic gases at absolute...

Two perfect monoatomic gases at absolute temperature `T_(1)` and `T_(2)` are mixed. There is no loss of energy. Find the temperature of the mixture if the number of moles in the gases are` n_(1)` and `n_(2)`.

Text Solution

Verified by Experts

The correct Answer is:
A, B
From energy conversation principle,
`E_i = E_f`
or `E_1 + E_2 = E`
`:. n_1(3/2RT_1) + n_2(3/2R T_2)= (n_1 + n_2)(3/2R T)_`
`:. T = (n_1 T_1 + n_2 T_2)/(n_1 + n_2)`.
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Knowledge Check

  • Two perfect gases at absolute temperature T_(1) and T_(2) are mixed. There is no loss of energy. The masses of the molecules are m_(1) and m_(2) . The number of molecules in the gases are n_(1) and n_(2) . The temperature of the mixture is

    A
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    B
    `T_(1) + T_(2)/2`
    C
    `n_(1)T_(1) + n_(2)T_(2)/n_(1) +n_(2)`
    D
    None of these

  • Two perifect gases at absolute temperatures T_(1) and T_(2) are mixed. The absolute temperature of the mixture is T. there is no loss of energy.If m_(1) and m_(2) are masses of molecules and n_(1) and n_(2) are number of molecules, then

    A
    `T=(T_(1)+T_(2))/(2)`
    B
    `T=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`
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    D
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  • Three perfect gases at absolute temperature T_(1), T_(2) and T_(3) are mixed. The masses f 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

    A
    `((T_(1) + T_(2) + T_(3)))/(3)`
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    D
    `(n_(1)^(2)T_(1)^(2) + n_(2)^(2)T_(2)^(2) + n_(3)^(2)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))`

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