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.)

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Heat & Thermodynamics

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{α ^{2}}{4} RJ / mol K

; then the value of

α

will be ____ . (Assume that the given diatomic gas has no vibrational mode.)

Answer: 3

Solution

For a mixture:

C_{v, mix} = \frac{n _{1} C _{v 1} + n _{2} C _{v 2}}{n _{1} + n _{2}}

. For monoatomic gas,

C_{v 1} = \frac{3 R}{2}

; for diatomic gas (no vibration),

C_{v 2} = \frac{5 R}{2}

. Thus

C_{v, mix} = \frac{1 \cdot \frac{3 R}{2} + 3 \cdot \frac{5 R}{2}}{1 + 3} = \frac{\frac{3 R + 15 R}{2}}{4} = \frac{9 R}{4}

. Given

C_{v, mix} = \frac{α ^{2}}{4} R

, we have

\frac{α ^{2}}{4} R = \frac{9 R}{4}

, so

α^{2} = 9

and

α = 3

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer

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