Two perfect gases at absolute temperatures $T_{1} a n d T_{2}$ are mixed. There is no loss of energy in this process. If $n_{1} a n d n_{2}$ are the respective number of molecules of the gases, the temperature of the mixture will be

Q. For the Paschen series the values of

n_{1}

and

n_{2}

in the expression

△ E = R_{H} \times c [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]

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At 327oC, the kinetic energy of a sample of N2 gas containing n1 molecules is 4E. At −123oC, the kinetic energy of another sample of N2 gas containing n2 molecules is E. Then:

The correct option is C n1=n2KE1KE2=n1T1n2T2T1=327oC=600K, T2=−123oC=K4EE=n1×NA×600n2×NA×150n1=n2

At $327^{o} C$ , the kinetic energy of a sample of $N_{2}$ gas containing $n_{1}$ molecules is $4 E$ . At $- 123^{o} C$ , the kinetic energy of another sample of $N_{2}$ gas containing $n_{2}$ molecules is $E$ . Then:

The correct option is C $n_{1} = n_{2}$
$\frac{K E_{1}}{K E_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
$T_{1} = 327^{o} C = 600 K$ , $T_{2} = - 123^{o} C = K$
$\frac{4 E}{E} = \frac{n_{1} \times N_{A} \times 600}{n_{2} \times N_{A} \times 150}$

$n_{1} = n_{2}$