Question

The specific heat at constant volume for the monoatomic argon is 0.075 kcal/kg-K whereas its gram molecular specific heat $C_{v} = 2.98$cal/mole/K. The mass of the argon atom is (Avogadro’s number $= 6.02 \times 10^{23}$molecules/mole)

• A. $6.60 \times 10^{- 23}gm$
• B. $3.30 \times 10^{- 23}gm$
• C. $2.20 \times 10^{- 23}gm$
• D. $13.20 \times 10^{- 23}gm$

Answer 1

Answer: (A)

$6.60 \times 10^{- 23}gm$

Answer 2

Molar specific heat = Molecular weight × Gram specific heat

$C_{v} = M \times c_{v}$

⇒$2.98\frac{calorie}{mole \times kelvin} = M \times 0.075\frac{kcal}{kg - kelvin} = M \times \frac{0.075 \times 10^{3}}{10^{3}}\frac{calorie}{gm \times kelvin}$

∴ molecular weight of argon $M = \frac{2.98}{0.075} = 39.7gm$

i.e. mass of $6.023 \times 10^{23}$atom = 39.7 gm ∴ mass of single atom $= \frac{39.7}{6.023 \times 10^{23}} = 6.60 \times 10^{- 23}gm$.