If one mole of a monoatomic gas - r1-53 - is mixed with one mole of a diatomic gas - r2-75 - the value of r for the mixture is -

The correct option is D 1.5For the mixture, r_mix=μ_1r_1r_1 1+μ_2r_2r_2 1μ_1r_1 1+μ_2r_2 1=1×53[ 53 1 ]+1×75[ 75 1 ][ 1[ 53 1 ] ]+[ 1[ 75 1 ] ] #160; ...

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Standard XII

Chemistry

Ideal Solutions

If one mole o...

Question

If one mole of a monoatomic gas $(\begin{matrix} r_{1} = \frac{5}{3} \end{matrix})$ is mixed with one mole of a diatomic gas $(\begin{matrix} r_{2} = \frac{7}{5} \end{matrix})$ , the value of r for the mixture is :

3.07

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1.5

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1.4

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1.53

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Solution

The correct option is D 1.5
For the mixture,
$r_{m i x} = \frac{\frac{μ_{1} r_{1}}{r_{1} - 1} + \frac{μ_{2} r_{2}}{r_{2} - 1}}{\frac{μ_{1}}{r_{1} - 1} + \frac{μ_{2}}{r_{2} - 1}} = \frac{\frac{1 \times \frac{5}{3}}{(\begin{matrix} \frac{5}{3} - 1 \end{matrix})} + \frac{1 \times \frac{7}{5}}{(\begin{matrix} \frac{7}{5} - 1 \end{matrix})}}{⎡ ⎢ ⎣ \begin{matrix} \frac{1}{(\begin{matrix} \frac{5}{3} - 1 \end{matrix})} \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} \frac{1}{(\begin{matrix} \frac{7}{5} - 1 \end{matrix})} \end{matrix} ⎤ ⎥ ⎦}$
$= \frac{3}{2} = 1.5$

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If one mole of a monoatomic gas (r1=53) is mixed with one mole of a diatomic gas (r2=75), the value of r for the mixture is :

The correct option is D 1.5For the mixture,rmix=μ1r1r1−1+μ2r2r2−1μ1r1−1+μ2r2−1=1×53(53−1)+1×75(75−1)⎡⎢⎣1(53−1)⎤⎥⎦+⎡⎢⎣1(75−1)⎤⎥⎦ =32=1.5

If one mole of a monoatomic gas $(\begin{matrix} r_{1} = \frac{5}{3} \end{matrix})$ is mixed with one mole of a diatomic gas $(\begin{matrix} r_{2} = \frac{7}{5} \end{matrix})$ , the value of r for the mixture is :