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 (integer)
3
Solution
<p>${C_V} = {f \over 2}R$</p>
<p>total degree of freedoms</p>
<p>$= 1 \times 3 + 3 \times 5 = 18$</p>
<p>${{{\alpha ^2}} \over 4} = {{18} \over {2n}} = {{18} \over {2 \times 4}}$</p>
<p>$\Rightarrow {\alpha ^2} = 9$</p>
<p>$\alpha = 3$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.