Question

The specific heat of Ar at constant volume is $0.075kcalk{g}^{-1}{K}^{-1}$. Calculate the atomic weight. $(R=2calmo{l}^{-1}{K}^{-1})$.

• A. $40$
• B. $40.4$
• C. $40.2$
• D. $40.8$

Answer 1

The correct option is A $40$

We know that the molar specific heat $C_v$ is given as the product of specific heat $S_v$ multiplied by atomic weight M. Thus
$M=\frac{C_v}{S_v}=\frac{\frac{3\times 2}{2}}{0.075}=40$