At constant volume, the specific heat of a gas is 0.025 $c a l K^{- 1} g^{- 1}$ and its molecular weight is 120 $g m o l^{- 1}$ . The atomicity of the gas is:
Given: R = 2 cal $K^{- 1} m o l^{- 1}$

Diatomic

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Triatomic

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Monoatomic

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None of the above

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Solution

The correct option is C Monoatomic
We know that molar heat capacity at constant volume,
$(C_{v}, m)$ = Specific heat at constant volume $\times$ Molecular weight.

= $0.025 \times 120 = 3 c a l K^{- 1} m o l^{- 1}$ ....(i)

Also, $C_{p} - C_{v} = R$
$\Rightarrow C_{p} = R + C_{v}$
= $C_{p} = 2 + 3 = 5 c a l K^{- 1} m o l^{- 1}$ .....(ii)

Now, $\frac{C_{p}}{C_{v}} = γ$

Using equation (i) and (ii)
$γ = \frac{5}{3} = 1.67$
This value shows that the gas is monoatomic.

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Similar questions

Q. At constant volume, the specific heat of a gas is 0.025

c a l K^{- 1} g^{- 1}

and its molecular weight is 120

g m o l^{- 1}

. The atomicity of the gas is:
Given: R = 2 cal

K^{- 1} m o l^{- 1}

Q. At constant volume, the specific heat of a gas is 0.025

c a l K^{- 1} g^{- 1}

and its molecular weight is 120

g m o l^{- 1}

. The atomicity of the gas is:
Given: R = 2 cal

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