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One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $\frac{\alpha^{2}}{4} \mathrm{R} \mathrm{J} / \mathrm{mol} \: \: \mathrm{K}$ ; then the value of $\alpha$ will be__________ (Assume that the given diatomic gas has no vibrational mode).
Option: 1
3

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

For monoatomic gas ,

$\mathrm{V_{1}=\frac{5}{3},n_{1}=1}$

For diatomic gas,

$\mathrm{V_{2}=\frac{7}{5},n_{2}=3}$

$\mathrm{U=U_{1}+U_{2}}$

$\mathrm{n_{\text {mix }}\left(c_v\right)_{\operatorname{mix}} T =n_1\left(c_v\right)_1 T+n_2\left(c_v\right)_2 T }$

$\mathrm{\frac{(\left.n_1+n_2\right) R}{v_{\operatorname{mix}}-1} =\frac{n_1 R}{v_1-1}+\frac{n_2 R}{v_2-1} }$

$\mathrm{\frac{4}{V_{\operatorname{mix}}-1} =\frac{1}{2 / 3}+\frac{3}{2 / 5} }$

$\mathrm{=\frac{3}{2}+\frac{15}{2} }$

$\mathrm{\frac{4}{v_{\text {mix }}-1} =9}$

$\mathrm{\frac{4}{9}+1 =V_{\text {mix }} }$

$\mathrm{v_{\text {mix }} =\frac{13}{9}}$

$\mathrm{\left ( C_{v} \right )_{mix}=\frac{R}{v_{mix}-1}=\frac{\alpha ^{2}}{4}R}$

$\mathrm{\frac{1}{\left ( \frac{13}{9}-1 \right )}=\frac{\alpha ^{2}}{4}}$

$\mathrm{\frac{9}{4}=\frac{\alpha ^{2}}{4}}$

$\mathrm{\alpha =3}$

The value of $\mathrm{\alpha}$ is 3

Posted by

rishi.raj

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