Question

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

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

Answer 1

Answer: (A)

Molar specific heat $=$ molecular weight $\times$ gram specific heat
$C_{v}=M \times c_{v}$
$2.98\, cal / mol - K =M \times 0.075\, kcal / kg - K$
$=M \times \frac{0.075 \times 10^{3}}{10^{3}} \,cal / g - K$
$\therefore $ Molecular weight of argon
$M=\frac{2.98}{0.075}=39.7\,g$
i.e., mass of $6.023 \times 10^{23}$ atom $=39.7 \,g$
Therefore, mass of single atom
$=\frac{39.7}{6.023 \times 10^{23}}=6.60 \times 10^{-23} g$