JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 23)
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).
Answer
3
Explanation
$${C_V} = {f \over 2}R$$
total degree of freedoms
$$ = 1 \times 3 + 3 \times 5 = 18$$
$${{{\alpha ^2}} \over 4} = {{18} \over {2n}} = {{18} \over {2 \times 4}}$$
$$ \Rightarrow {\alpha ^2} = 9$$
$$\alpha = 3$$
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